## 1. The slope of line and angles between two lines

Chapter 10

Straight Lines

The slope of line and angles between two lines:

A line is sometimes called a straight line or, more archaically, a right line (Casey 1893), to emphasize that it has no "wiggles" anywhere along its length. While lines are intrinsically one-dimensional objects, they may be embedded in higher dimensional spaces. Harary (1994) called an edge of a graph a "line."

A straight line is just a line with no curves. So, a line that extends to both sides till infinity and has no curves is called a straight line.

The two properties of straight lines in Euclidean geometry are that they have only one dimension, length, and they extend in two directions forever.

Note: If the area of the triangle ABC is zero, then three points A, B and C lie on

a line, i.e., they are collinear.

Slope of a Line:

A line in a coordinate plane forms two angles with the x-axis, which are supplementary.

The angle (say) q made by the line l with positive direction of x-axis and measured anti clockwise is called the inclination of the line. Obviously £ q £ 180°

lines parallel to x-axis, or coinciding with x-axis, have inclination of 0°. The inclination of a vertical line (parallel to or coinciding with y-axis) is 90°.

Definition: If q is the inclination of a line l, then tan q is called the slope or gradient of

the line l. The slope of a line whose inclination is 90° is not defined.

The slope of a line is denoted by m.

Thus, m = tan q, q ¹ 90°

It may be observed that the slope of x-axis is zero and slope of y-axis is not defined.

Slope of a line when coordinates of any two points on the line are given:

Let P(x1, y1) and Q(x2, y2) be two points on non-vertical line l whose inclination is q.

Case I When angle q is acute:

ÐMPQ = q. ... (1)

Therefore, slope of line l = m = tan q.

Case II When angle q is obtuse:

we have ÐMPQ = 180° – q.

Therefore, q = 180° – ÐMPQ.

Now, slope of the line l

m = tan q

= tan ( 180° – ÐMPQ) = – tan ÐMPQ

Example : Find the slope of the lines:

(a) Passing through the points (3, – 2) and (–1, 4),

(c) Passing through the points (3, – 2) and (3, 4),

(c) Making inclination of 30° with the positive direction of x-axis.

Solution: (a) The slope of the line through (3, – 2) and (– 1, 4) is

m = (4 – (-2)) /(-1 – 3) = 6 / - 4 = -3/2

(b) The slope of the line through the points (3, – 2) and (3, 4) is

m  = (4- (- 2)) / (3-3) = 6 / 0 , which is not defined.

(c) Here inclination of the line a = 60°. Therefore, slope of the line is

m = tan 30° = 1 / √3

## 1. Sections of a cone: parabola, hyperbola, ellipse

Chapter 11

Conic Sections

Sections of a cone: circle, parabola, hyperbola, ellipse:

When the plane cuts the nappe (other than the vertex) of the cone, we have the following situations:

(a) When b = 90o, the section is a circle .

(b) When a < b < 90o, the section is an ellipse .

(c) When b = a; the section is a parabola .

(In each of the above three situations, the plane cuts entirely across one nappe of the cone).

(d) When 0 £ b < a; the plane cuts through both the nappes and the curves of intersection is a hyperbola .

Circle: Set of points in a plane equidistant from a fixed point. A circle with radius r and centre (h, k) can be represented as (x – h)+ (y – k)= r2

Parabola: Set of points in a plane that are equidistant from a fixed-line and point. A parabola with a > 0, focus at (a, 0), and directrix x = – a can be represented as y= 4ax

In parabola y= 4ax, the length of the latus rectum is given by 4a.

Ellipse: The sum of distances of a set of points in a plane from two fixed points is constant. An ellipse with foci on the x-axis can be represented as:

Hyperbola: The difference of distances of set of points in a plane from two fixed points is constant. The hyperbola with foci on the x-axis can be represented as:

## 1. Coordinate axes and planes in 3D

Chapter 12

Introduction to Three-Dimensional Geometry

Coordinate axes and planes in 3D:

Let Consider three planes intersecting at a point O such that these three planes are mutually perpendicular to each other.

These three planes intersect along the lines XOX, YOY and ZOZ, called the x, y and z-axes, respectively.

The point O is called the origin of the coordinate system. The three coordinate planes divide the space into eight parts known as octants. These octants could be named as XOYZ,  XOYZ, XOYZ,  XOYZ, XOYZ,  XOYZ,  XOYZ and XOYZ. and denoted by I, II, III,.. ..., VIII , respectively.

Coordinates of a Point in Space

Let OA = x, OB = y and OC = z. Then, the point P will have the coordinates x, y and z and we write P (x, y, z). Conversely, given x, y and z, we locate the three points A, B and C on the three coordinate axes. Through the points A, B and C we draw planes parallel to the YZ-plane, ZX-plane and XY-plane,respectively.

In three-dimensional geometry, the coordinates of a point P is always written in the form of P(x, y, z), where x, y and z are the distances of the point, from the YZ, ZX and XY-planes.

• The coordinates of any point at the origin is (0,0,0)
• The coordinates of any point on the x-axis is in the form of (x,0,0)
• The coordinates of any point on the y-axis is in the form of (0,y,0)
• The coordinates of any point on the z-axis is in the form of (0,0,z)
• The coordinates of any point on the XY-plane is in the form (x, y, 0)
• The coordinates of any point on the YZ-plane is in the form (0, y, z)
• The coordinates of any point on the ZX-plane is in the form (x, 0, z)

Sign of Coordinates in Different Octants:

The sign (+ or -) of the coordinates of a point determines the octant in which the point lies.

Example: In Fig 12.3, if P is (2,4,5), find the coordinates of F.

Solution For the point F, the distance measured along OY is zero. Therefore, the

coordinates of F are (2,0,5).

Example  Find the octant in which the points (–3,1,2) and (–3,1,– 2) lie.

Solution From the Table 12.1, the point (–3,1, 2) lies in second octant and the point

(–3, 1, – 2) lies in octant VI.

## 2. Conditions for parallelism and perpendicularity of lines and Collinearity of three points

Conditions for parallelism and perpendicularity of lines and Collinearity of three points:

In a coordinate plane, suppose that non-vertical lines l1 and l2 have slopes m1

and m2, respectively. Let their inclinations be a and b, respectively.

If the line l1 is parallel to l2 (Fig 10.4), then their

inclinations are equal, i.e.,a = b, and hence, tan a = tan b

Therefore m1 = m2, i.e., their slopes are equal.

Conversely, if the slope of two lines l1 and l2 is same,

i.e., m1 = m2.

Then   tan a = tan b.

By the property of tangent function (between 0° and 180°), a = b.

Therefore, the lines are parallel.

Hence, two non vertical lines l1 and l2 are parallel if and only if their slopes are equal.

If the lines l1 and l2 are perpendicular , then b = a + 90°.

Therefore, tan b = tan (a + 90°)

= – cot a = -1/ tan a

i.e., m1= -1/m2   or m1 m2 = -1

Conversely, if m1 m2 = – 1, i.e., tan a tan b = – 1.

Then tan a = – cot b = tan (b + 90°) or tan (b – 90°)

Therefore, a and b differ by 90°.

Thus, lines l1 and l2 are perpendicular to each other.

Hence, two non-vertical lines are perpendicular to each other if and only if their slopes are negative reciprocals of each other,

Example:  Line through the points (–2, 6) and (4, 8) is perpendicular to the line

through the points (8, 12) and (x, 24). Find the value of x.

Solution: Slope of the line through the points (– 2, 6) and (4, 8) is

m = (8 – 6) /(4  – (-2)) = 2 / 6 = 1/ 3

Slope of the line through the points (8, 12) and (x, 24) is

m = (24 – 12) /(x – 8) = 12 / (x-8)

Since two lines are perpendicular,

m1 m2 = –1, which gives

• [12 / (x-8)] . [1 / 3] = - 1
• 4 / ( x – 8 ) = - 1
•  x – 8 = - 4
• x = 4

Angle between two lines

Let L1 and L2 be two non-vertical lines with slopes m1 and m2, respectively. If a1 and a2 are the inclinations of lines L1 and L2, respectively. Then m1= tan a1 and m2=tana2

Let q and f be the adjacent angles between the lines L1 and L2 . Then q = α2 - a1  and a1, a2 ¹ 90o

Therefore tan q = tan (a2 – a1) =[ tana2 – tana1] / [1+ tana2 –tana1] = [m2-m1] / [1+ m1m2]

(as 1 + m1m2 ¹ 0)

and  f = 180° – q so that

tan f = tan (180° – q ) = – tan q = - [m2-m1] / [1+ m1m2] (as 1 + m1m2 ¹ 0)

Now, there arise two cases:

which means that q will be obtuse and f will be acute.

Thus, the acute angle (say q) between lines L1 and L2 with slopes m1 and m2,

respectively, is given by

……………………(1)

The obtuse angle (say f) can be found by using  f =1800q.

Example: If P (-2, 1), Q (2, 3) and R (-2, -4) are three points, find the angle between the straight lines PQ and QR.

The slope of PQ is given by

m = ( y2 – y1 ) / (x2 – x1)

m =( 3 – 1 ) / (2 – (-2 ))

m= 2/4

Therefore, m1=1/2

The slope of QR is given by

m= (−4−3) / (−2−2)

m= 7/4

Therefore, m2 = 7/4

Substituting the values of m2 and m1 in the formula for the angle between two lines when we know the slopes of two sides, we have,

tan θ=± (m2 – m1 ) / (1+m1m2)

tan θ=± ((7/4) – (1/2) ) / (1+ (1/2)(7/4))

tan θ=± (2/3)

Therefore,  θ = tan -1 (⅔)

Co linearity of three points

We know that slopes of two parallel lines are equal. If two lines having the same slope pass through a common point, then two lines will coincide. Hence, if A, B and C are three points in the XY-plane, then they will lie on a line, i.e., three points are collinear  if and only if

slope of AB = slope of BC.

Example: Three points P (h, k), Q (x1, y1) and R (x2, y2) lie on a line. Show that

(h x1) (y2 – y1) = (k y1) (x2 – x1).

Solution : Since points P, Q and R are collinear, we have

Slope of PQ = Slope of QR,

or   (h x1) (y2 – y1) = (k y1) (x2 – x1).

## 3. Various Forms of the Equation of a Line

Various Forms of the Equation of a Line:

Different forms of equations of a straight line

• Point-slope form equation of line. Consider a non-vertical line L whose slope is m, A(x,y) be an arbitrary point on the line and. ...
• Two-point form equation of line. ...
• Slope-intercept form equation of line. ...
• Intercept form. ...
• Normal form.

Horizontal and vertical lines:

If a horizontal line L is at a distance a from thex-axis then ordinate of every point lying on the line is either a or – a .

Therefore, equation of the line L is either y = a or y = a. Choice of sign will depend upon the position of the line according as the line is above or below the y-axis.

Similarly,

the equation of a vertical line at a distance b from the y-axis is either x = b or x = – b

Example 6 Find the equations of the lines parallel to axes and passing through (– 2, 3).

Solution Position of the lines is shown in the Figure. The y-coordinate of every point on the line parallel to x-axis is 3, therefore, equation of the line parallel tox-axis and passing through

(– 2, 3) is y = 3. Similarly, equation of the line parallel to y-axis and passing through (– 2, 3)

is x = – 2.

Point-slope form

Suppose that P0 (x0, y0) is a fixed point on a non-vertical line L, whose slope is m. Let P (x, y) be an arbitrary point on L  .

Then, by the definition, the slope of L is given by

is the equation of line.

Example : Find the equation of the line through (– 2, 3) with slope – 4.

Solution: Here m = – 4 and given point (x0 , y0) is (– 2, 3).

By slope-intercept form above formula , equation of the given line is

y – 3 = – 4 (x + 2) or

4x + y + 5 = 0, which is the required equation.

Two-point form

Let the line L passes through two given points P1 (x1, y1) and P2 (x2, y2).

Let P (x, y) be a general point on L . The three points P1, P2 and P are collinear, therefore, we have slope of P1P = slope of P1P2

is the equation of the line passing through the points (x1, y1) and (x2, y2).

Example: Write the equation of the line through the points (1, –1) and (3, 5).

Solution : Here x1 = 1, y1 = – 1, x2 = 3 and y2 = 5. Using two-point form  above formula

for the equation of the line, we have

• –3x + y + 4 = 0, which is the required equation.

Slope-intercept form Sometimes a line is known to us with its slope and an

intercept on one of the axes. We will now find equations of such lines.

Case I Suppose a line L with slope m cuts the y-axis at a distance c from the origin

. The distance c is called the y-intercept of the line L. Obviously, coordinates of the point where the line meet the y-axis are (0, c). Thus, L has slope m and passes through a fixed point (0, c).

Therefore, by point-slope form, the equation of L is

y - c = m( x - 0 ) or y = mx + c

Thus, the point (x, y) on the line with slope m and y-intercept c lies on the line if and only if

y = mx +c ......................(3)

Note that the value of c will be positive or negative according as the intercept is made

on the positive or negative side of the y-axis, respectively.

Case II Suppose line L with slope m makes x-intercept d. Then equation of L is

y = m(x – d) …………………... (4)

Students may derive this equation themselves by the same method as in Case I.

Intercept – form

Suppose a line L makes x-intercept a and y-intercept b on theaxes. Obviously L meets x-axis at the point (a, 0) and y-axis at the point (0, b) . By two-point form of the equation of the line, we have,

y – 0 =   b-00-a( x – a)

• ay = – bx + ab
• xa  + yb  = 1

is equation of the line making intercepts a and b on x-and y-axis, respectively

Normal form

Suppose a non-vertical line is known to us with following data:

(i) Length of the perpendicular (normal) from origin to the line.

(ii) Angle which normal makes with the positive direction of x-axis.

Let L be the line, whose perpendicular distance from origin O be OA = p and the

angle between the positive x-axis and OA be ÐXOA = w.

Draw perpendicular AM on the x-axis in each case.

In each case, we have OM = p cos w and MA = p sin w, so that the coordinates of the

point A are (p cos w, p sin w).

Further, line L is perpendicular to OA. Therefore

The slope of the line L

the equation of the line L at point A(pcos w, psin w ) is

x cos w + y sin w = p.

Hence, the equation of the line having normal distance p from the origin and angle w

## 4. General Equation of a Line

General Equation of a Line

General equation of a line is Ax + By + C = 0. The inclination of angle θ to a line with a positive direction of X-axis in the anti-clockwise direction, the tangent of angle θ is said to be slope or gradient of the line and is denoted by m

General equation of first degree in two variables,

Ax + By + C = 0, where A, B and C are real constants such that A and B are not zero

simultaneously. Graph of the equation Ax + By + C = 0 is always a straight line.

Therefore, any equation of the form Ax + By + C = 0, where A and B are not zero

simultaneously is called general linear equation or general equation of a line.

Different forms of Ax + By + C = 0 The general equation of a line can be

reduced into various forms of the equation of a line.

If B = 0, then x = - C/A which is a vertical line whose slope is undefined and x-intercept  is  -C/A.

If C = 0, then Ax + By + C = 0 can be written as Ax + By = 0, which is a line passing through the origin and, therefore, has zero intercepts on the axes.

(Q1) The equation of a line is given by, 2x – 6y +3 = 0. Find the slope and both the intercepts.
Solution:

The given equation 2x – 6y + 3 = 0 can be represented in slope-intercept form as:

y = x/3 + 1/2

Comparing it with y = mx + c,
Slope of the line, m = 1/3

Also, the above equation can be re-framed in intercept form as;

x/a + y/b = 1

2x – 6y = -3

x/(-3/2) – y/(-1/2) = 1

Thus, x-intercept is given as a = -3/2 and y-intercept as b = 1/2.

(Q2) The equation of a line is given by, 13x – y + 12 = 0. Find the slope and both the intercepts.

Solution: The given equation 13x – y + 12 = 0 can be represented in slope-intercept form as:

y = 13x + 12
Comparing it with y = mx + c,
Slope of the line, m = 13

Also, the above equation can be re-framed in intercept form as;

x/a + y/b = 1

13x – y = -12

x/(-12/13) + y/12 = 0

Thus, x-intercept is given as a = -12/13 and y-intercept as b = 12.

Example: Show that two lines a1x + b1 y + c1 = 0 and a2x + b2 y + c2 = 0, where b1,b2 ¹ 0 are:

## 5. Distance of a Point From a Line and Distance between two parallel lines

Distance of a Point from a Line and Distance between two parallel lines

The distance of a point from a line is the length of the perpendicular drawn from the point to the line. Let L : Ax + By + C = 0 be a line, whose distance from the point P (x1, y1) is d. Draw a perpendicular PM from the point P to the line L

Thus, the perpendicular distance (d) of a line Ax + By+ C = 0 from a point (x1, y1) is given by

Distance between two parallel lines

We know that slopes of two parallel lines are equal. Therefore, two parallel lines can be taken in the form

y = mx + c1 ... (1)

and

y = mx + c2 ... (2)

Line (1) will intersect x-axis at the point A(-c/m,0)

Distance between two lines is equal to the length of the perpendicular from point A to line (2). Therefore, distance between the lines (1) and (2) is

If lines are given in general form, i.e., Ax + By + C1 = 0 and Ax + By + C2 = 0,

then above formula will take the form

Example:

Example: If the lines 2x + y - 3 = 0, 5x + ky - 3 = 0 and 3x - y - 2 = 0 are concurrent, find the value of k.

Solution Three lines are said to be concurrent, if they pass through a common point,

i.e., point of intersection of any two lines lies on the third line. Here given lines are

2x + y – 3 = 0 ... (1)

5x + ky – 3 = 0 ... (2)

3x y – 2 = 0 ... (3)

Solving (1) and (3) by cross-multiplication method, we get

or  x= 1, y= 1

Therefore, the point of intersection of two lines is (1, 1). Since above three lines are

concurrent, the point (1, 1) will satisfy equation (2) so that

5.1 + k .1 – 3 = 0 or k = – 2.

## 2. Standard equations, Properties and Application of a circle

Standard equations , Properties and Application of a circle

Circle:

Definition: A circle is the set of all points in a plane that are equidistant from a fixed

point in the plane.

The fixed point is called the centre of the circle and the distance from the centre

to a point on the circle is called the radius of the circle.

Given C (h, k) be the centre and r the radius of circle. Let P(x, y) be any point on

the circle . Then, by the definition, | CP | = r . By the distance formula,

we have

(x-h)2 + (y-k)2= r2

This is the required equation of the circle with centre at (h,k) and radius r .

General form of Equation of a Circle

The general equation of any type of circle is represented by:

x2 + y2 + 2gx + 2fy + c = 0, for all values of g, f and c.

Adding g2 + f2 on both sides of the equation gives,

x2 + 2gx + g2+ y2 + 2fy + f2= g2 + f2 − c ………………(1)

Since, (x+g)= x2+ 2gx + g2 and (y+f)=y+ 2fy + f2 substituting the values in equation (1), we have

(x+g)2+ (y+f)= g+ f2−c …………….(2)

Comparing (2) with (x−h)+ (y−k)= a2, where (h, k) is the center and ‘a’ is the radius of the circle.

h=−g, k=−f

a2 = g2+ f2−c

Therefore,

x+ y+ 2gx + 2fy + c = 0, represents the circle with centre (−g,−f) and radius equal to a2 = g+ f2− c.

• If g+ f> c, then the radius of the circle is real.
• If g+ f= c, then the radius of the circle is zero which tells us that the circle is a point that coincides with the center. Such a type of circle is called a point circle
• g+ f<c, then the radius of the circle become imaginary. Therefore, it is a circle having a real center and imaginary radius.

N.B.:  Standard Equation of Circle:

centre (0, 0) and Radius (r)

Equation of circle in centre radius form:

Centre (h, k), Radius = r

Equation of circle in General form:

Where (–g, –f ) centre

r2 = g2 + f2 – c .

Equation of circle with points P(x1, y1) and Q(x2, y2) as extremities of diameter is

(x – x1) (x – x2) + (y – y1) (y – y2) = 0

Equation of circle through three non-collinear points P(x1, y1), Q(x2, y2) and R(x3, y3) is

Area of circle = πr2

Perimeter = 2πr, where r is the radius.

Example: Find an equation of the circle with centre at (0,0) and radius r.

Solution :Here h = k = 0. Therefore, the equation of the circle is x2 + y2 = r2

Example: Find the equation of the circle with centre (–3, 2) and radius 4.

Solution: Here h = –3, k = 2 and r = 4. Therefore, the equation of the required circle is

(x + 3) 2 + (y –2)2 = 16

Example  : Find the centre and the radius of the circle x2 + y2 + 8x + 10y – 8 = 0

Solution : The given equation is

(x2 + 8x) + (y22+ 10y) = 8

Now, completing the squares within the parenthesis, we get

(x2+ 8x + 16) + (y2 + 10y + 25) = 8 + 16 + 25

i.e. (x + 4)2 + (y + 5)2 = 49

i.e. {x – (– 4)} 2+ {y – (–5)} 2 = 72

Therefore, the given circle has centre at (– 4, –5) and radius 7.

Example : Find the equation of the circle which passes through the points (2, – 2), and (3, 4) and whose centre lies on the line x + y = 2.

Solution: Let the equation of the circle be (x – h)2 + (y – k)2 = r2.

Given that the circle passes through the points (2, –2) and (3, 4).

Thus,

(2 – h)2 + (–2 – k)2 = r2….(1)

and (3 – h)2 + (4 – k)2 = r2….(2)

Also, given that the centre lies on the line x + y = 2.

h + k = 2 ….(3)

Solving the equations (1), (2) and (3), we get

h = 0.7, k = 1.3 and r2 = 12.58

Hence, the equation of the required circle is

(x – 0.7)2 + (y – 1.3)2 = 12.58

## 3. Standard equations, Properties and Application of a parabola

Standard equations , Properties and Application of a parabola

Parabola:

Definition: A parabola is the set of all points in a plane that are equidistant from a fixed line

and a fixed point (not on the line) in the plane.

The fixed line is called the directrix of the parabola and the fixed point F is called the focus 11. (‘Para’ means ‘for’ and ‘bola’ means ‘throwing’, i.e., the shape described when you throw a ball in the air).

A line through the focus and perpendicular to the directrix is called the axis of the parabola.

The point of intersection of parabola with the axis is called the vertex of the parabola.

Standard equations of parabola

The equation of a parabola is simplest if the vertex is at the origin and the axis of symmetry is along the x-axis or y-axis. The four possible such orientations of parabola.

Let P(x, y) be any point on the parabola such that

PF = PB, ... (1)

where PB is perpendicular to l. The coordinates of B are (– a, y). By the distance

formula, we have

(PF) 2 = (x – a)2 + y2  and (PB) 2 =  (x + a) 2

Since PF = PB, we have

(x – a) 2 + y2 = (x + a) 2

or x2 – 2ax + a2 + y2 = x2 + 2ax + a2

or y2 = 4ax ( a > 0). is  Standard equation of Parabola.

Latus rectum

Definition : Latus rectum of a parabola is a line segment perpendicular to the axis of

the parabola, through the focus and whose end points lie on the parabola .

To find the Length of the latus rectum of the parabola y2 = 4ax .

By the definition of the parabola, AF = AC.

But AC = FM = 2a

Hence AF = 2a.

And since the parabola is symmetric with respect to x-axis AF = FB and so

AB = Length of the latus rectum = 4a.

Example : Find the coordinates of the focus, axis, the equation of the directrix and latus rectum of the parabola y2 = 8x.

Solution The given equation involves y2, so the axis of symmetry is along the x-axis.

The coefficient of x is positive so the parabola opens to the right. Comparing with the given equation y2 = 4ax, we find that a = 2.

Thus, the focus of the parabola is (2, 0) and the equation of the directrix of the parabola is

x = – 2 .

Length of the latus rectum is 4a = 4 × 2 = 8.

Properties:

1. Parabola is symmetric with respect to the axis of the parabola.If the equation

has a y2 term, then the axis of symmetry is along the x-axis and if the

equation has an x2 term, then the axis of symmetry is along the y-axis.

2. When the axis of symmetry is along the x-axis the parabola opens to the

(a) right if the coefficient of x is positive,

(b) left if the coefficient of x is negative.

3. When the axis of symmetry is along the y-axis the parabola opens

(c) upwards if the coefficient of y is positive.

(d) downwards if the coefficient of y is negative

Example: Find the equation of the parabola with focus (2,0) and directrix x = – 2.

Solution :Since the focus (2,0) lies on the x-axis, the x-axis itself is the axis of the

parabola. Hence the equation of the parabola is of the form either

y2 = 4ax

• y2 = – 4ax. Since the directrix is x = – 2 and the focus is (2,0), the parabola

is to be of the form y2 = 4ax with a = 2.

Hence the required equation is

y2 = 4(2)x = 8x

Example: Find the equation of the parabola with vertex at (0, 0) and focus at (0, 2).

Solution: Since the vertex is at (0,0) and the focus is at (0,2) which lies on y-axis, the

y-axis is the axis of the parabola. Therefore, equation of the parabola is of the form

x2 = 4ay. thus, we have

x2 = 4(2)y, i.e., x2= 8y.

## 2. Distance between 2 points, section formula

Distance between 2 points, section formula

Distance between Two Points

If P (x1, y1, z1) and Q (x2, y2, z2) are the two points, then the distance between P and Q is given by:

Now PA = y2 – y1, AN = x2 – x1 and NQ = z2 – z1

Hence PQ2 = (x2 – x1)2 + (y2 – y1) 2+ (z2 – z1) 2

Therefore PQ =  [(x2 – x1)2 + (y2 – y1) 2+ (z2 – z1) 2 ]This gives us the distance between two points (x1, y1, z1) and (x2, y2, z2).

In particular, if x1 = y1 = z1 = 0, i.e., point P is origin O, then OQ =  √(x22 + y22 + z22)

which gives the distance between the origin O and any point Q (x2, y2, z2).

PQ=(x2-x1)2+(y2-y1)2+(z2-z1)2

Section Formula

Example : Find the equation of set of points P such that PA2 + PB2 = 2k2, where A and B are the points (3, 4, 5) and (–1, 3, –7), respectively.

Solution Let the coordinates of point P be (x, y, z).

Here PA2 = (x – 3) 2 + (y – 4) 2 + ( z – 5) 2

PB2 = (x + 1) 2 + (y – 3) 2 + (z + 7) 2

By the given condition PA2 + PB2 = 2k2, we have

(x – 3) 2 + (y – 4) 2 + (z – 5) 2 + (x + 1) 2 + (y – 3) 2 + (z + 7) 2 = 2k2

i.e., 2x2 + 2y2 + 2z2 – 4x – 14y + 4z = 2k2 – 109.

Let the two given points be P(x1, y1, z1) and Q (x2, y2, z2). Let the point R (x, y, z) divide PQ in the given ratio m : n internally

To determine the coordinates of the point P, the following steps are followed:

• Draw AL, PN, and BM perpendicular to XY plane such that AL || PN || BM as shown above.
• The points L, M and N lie on the straight line formed due to the intersection of a plane containing AL, PN and BM and XY- plane.
• From point P, a line segment ST is drawn such that it is parallel to LM.
• ST intersects AL externally at S, and it intersects BM at T internally.

Since ST is parallel to LM and AL || PN || BM, therefore, the quadrilaterals LNPS and NMTP qualify as parallelograms.

Also, ∆ASP ~∆BTP therefore,

Rearranging the above equation we get,

Sectional Formula (Internally)

Thus, the coordinates of the point P(x, y, z) dividing the line segment joining the points A(x1, y1, z1) and B(x2, y2, z2) in the ratio m:n internally are given by:

Sectional Formula (Externally)

If the given point P divides the line segment joining the points A(x1, y1, z1) and B(x2, y2, z2) externally in the ratio m:n, then the coordinates of P are given by replacing n with –n as:

This represents the section formula for three dimension geometry.

If the point P divides the line segment joining points A and B internally in the ratio k:1, then the coordinates of point P will be

What if the point P dividing the line segment joining points A(x1, y1, z1) and B(x2, y2, z2) is the midpoint of line segment AB?

In that case, if P is the midpoint, then P divides the line segment AB in the ratio 1:1, i.e. m=n=1. Coordinates of point P will be given as:

Therefore, the coordinates of the midpoint of line segment joining points A(x1, y1, z1) and B(x2, y2, z2) are given by,

If the point R divides PQ externally in the ratio m : n, then its coordinates are obtained by replacing n by – n so that coordinates of point R will be

The above procedure can be repeated by drawing perpendiculars to XZ and YZ- planes to get the x and y coordinates of the point P that divides the line segment AB in the ratio m:n internally.

Sectional Formula (Internally)

Thus, the coordinates of the point P(x, y, z) dividing the line segment joining the points A(x1, y1, z1) and B(x2, y2, z2) in the ratio m:n internally are given by:

Sectional Formula (Externally)

If the given point P divides the line segment joining the points A(x1, y1, z1) and B(x2, y2, z2) externally in the ratio m:n, then the coordinates of P are given by replacing n with –n as:

This represents the section formula for three dimension geometry.

If the point P divides the line segment joining points A and B internally in the ratio k:1, then the coordinates of point P will be

What if the point P dividing the line segment joining points A(x1, y1, z1) and B(x2, y2, z2) is the midpoint of line segment AB?

In that case, if P is the midpoint, then P divides the line segment AB in the ratio 1:1, i.e. m=n=1.

Coordinates of point P will be given as:

Therefore, the coordinates of the midpoint of line segment joining points A(x1, y1, z1) and B(x2, y2, z2) are given by,

If the point R divides PQ externally in the ratio m : n, then its coordinates are obtained by replacing n by – n so that coordinates of point R will be

## 3. Coordinates of a points

Coordinates of a points

Having chosen a fixed 3D coordinate system in the space, we can associate a given point in the space using three coordinates (x,y,z) and conversely, given a triplet of three numbers (x, y, z), how, we locate a point in the space.

Locate the point “x” on the X-axis. From the point x, moving parallel to the Y-axis, locate the point “y”. Similarly, from the determined point, moving parallel to the Z-axis, locate the point “z”. This is the final coordinate point in the three-dimensional plane, which we are looking for.

when the co-ordinates of the point are given, then we have to draw three planes parallel to XY, YZ and ZX plane meeting the three axes in points A, B and C as shown in the figure. Let OA = x , OB = y and OC = z. Then the coordinates of the point are given as (x,y,z).

The above the co-ordinates of P are given by (x, y, z). The coordinates of the origin O is (0, 0, 0) Also the coordinates of the point A is given by (x, 0, 0)as A lies completely on the x-axis. Similarly, the coordinates of any point on y-axis is given as (0, y, 0) and on the z-axis, the coordinates are given as (0, 0, z). Also the coordinates of any point in three planes XY, YZ and ZX will be (x,y,0), (0,y,z) and (x,0,z) respectively.

Example: The coordinates of a point are (3, -2, 5). Write down the coordinates of seven points such that the absolute values of their coordinates are the same as those of the coordinates of the given point.

Solution:

Given:

Point (3, -2, 5)

The Absolute value of any point(x, y, z) is given by,

√(x2 + y2 + z2)

We need to make sure that absolute value to be the same for all points.

So let the point A(3, -2, 5)

Remaining 7 points are:

Point B(3, 2, 5) (By changing the sign of y coordinate)

Point C(-3, -2, 5) (By changing the sign of x coordinate)

Point D(3, -2, -5) (By changing the sign of z coordinate)

Point E(-3, 2, 5) (By changing the sign of x and y coordinate)

Point F(3, 2, -5) (By changing the sign of y and z coordinate)

Point G(-3, -2, -5) (By changing the sign of x and z coordinate)

Point H(-3, 2, -5) (By changing the sign of x, y and z coordinate)

Example: Find the ratio in which the line joining (2, 4, 5) and (3, 5, 4) is divided by the yz-plane.

Solution:

Given:

The points (2, 4, 5) and (3, 5, 4)

By using the section formula,

We know X coordinate is always 0 on yz-plane

So, let Point C(0, y, z), and let C divide AB in ratio k: 1

Then, m = k and n = 1

A(2, 4, 5) and B(3, 5, 4)

The coordinates of C are:

On comparing we get,

[3k + 2] / [k + 1] = 0

3k + 2 = 0(k + 1)

3k + 2 = 0

3k = – 2

k = -2/3

We can say that, C divides AB externally in ratio 2: 3

Question 1:

Find the equation of the set of points which are equidistant from the points (1, 2, 3) and (3, 2, –1).

Solution:

Assume that P (x, y, z) be the point that is equidistant from two points A(1, 2, 3) and B(3, 2, –1).

Thus, we can say that, PA = PB

Take square on both the sides, we get

PA= PB2

It means that,

(x-1)2 + (y-2)2+(z-3)2 = (x-3)2+(y-2)2+(z+1)2

x2 – 2x + 1 + y2 – 4y + 4 + z2 – 6z + 9 = x2 – 6x + 9 + y2 – 4y + 4 + z2 + 2z + 1

Now, simplify the above equation, we get:

–2x –4y – 6z + 14 = –6x – 4y + 2z + 14

– 2x – 6z + 6x – 2z = 0

4x – 8z = 0

x – 2z = 0

Hence, the required equation for the set of points is x – 2z = 0.

Question 2:

Given that P (3, 2, –4), Q (5, 4, –6) and R (9, 8, –10) are collinear. Find the ratio in which Q

divides PR.

Solution:

Assume that the point Q (5, 4, –6) divide the line segment joining points P (3, 2, –4) and R (9, 8, –10) in the ratio k:1.

Therefore, by using the section formula, we can write it as:

(5, 4, -6) = [ (k(9)+3)/(k+1), (k(8)+2)/(k+1), (k(-10)-4)/(k+1)]

(9k+3)/(k+1) = 5

Now, bring the L.H.S denominator to the R.H.S and multiply it

9k+3 = 5k+5

Now, simplify the equation to find the value of k.

4k= 2

k = 2/4

k=½

Therefore, the value of k is ½.

Hence, the point Q divides PR in the ratio of 1:2

Question 3:

Prove that the points: (0, 7, 10), (–1, 6, 6) and (–4, 9, 6) are the vertices of a right-angled triangle

Solution:

Let the given points be A = (0, 7, 10), B = (–1, 6, 6), and C = (–4, 9, 6)

Now, find the distance between the points

Finding for AB:

AB = √ [(-1-0)2 + (6-7)+(6-10)2]

AB = √ [(-1)2 + (-1)+(-4)2]

AB = √(1+1+16)

AB = √18

AB = 3√2 …. (1)

Finding for BC:

BC= √ [(-4+1)2 + (9-6)+(6-6)2]

BC = √ [(-3)2 + (3)+(-0)2]

BC = √(9+9)

BC = √18

BC = 3√2 …..(2)

Finding for CA:

CA= √ [(0+4)2 + (7-9)+(10-6)2]

CA = √ [(4)2 + (-2)+(4)2]

CA = √(16+4+16)

CA = √36

CA = 6 …..(3)

Now, by Pythagoras theorem,

AC2 = AB2 + BC…..(4)

Now, substitute (1),(2), and (3) in (4), we get:

6= ( 3√2)2 + ( 3√2)2

36 = 18+18

36 = 36

The given points obey the condition of Pythagoras Theorem.

Hence, the given points are the vertices of a right-angled triangle.

Question 4:

Calculate the perpendicular distance of the point P(6, 7, 8) from the XY – Plane.

(a)8  (b)7   (c)6  (d) None of the above

Solution:

A correct answer is an option (A)

Explanation:

Assume that A be the foot of perpendicular drawn from the point P (6, 7, 8) to the XY plane and the distance of this foot A from P is z-coordinate of P, i.e., 8 units

Hence, the correct answer is an option (a)

Question 5:

If a parallelopiped is formed by planes drawn through the points (2, 3, 5) and (5, 9, 7) parallel to the coordinate planes, then find the length of edges of a parallelopiped and length of the diagonal

Solution:

Let A = (2, 3, 5), B = (5, 9, 7)

To find the length of the edges of a parallelopiped = 5 – 2, 9 – 3, 7 – 5

It means that 3, 6, 2.

Now, to find the length of a diagonal = √(32 + 62 + 22)

= √(9+36+4)

= √49

= 7

Therefore, the length of a diagonal of a parallelopiped is 7 units.

## 4. Standard equations, Properties and Application of a ellipse

Standard equations , Properties and Application of a ellipse

Ellipse

Definition: An ellipse is the set of all points in a plane, the sum of whose distances from two fixed points in the plane is a constant. The two fixed points are called the foci (plural of ‘focus’) of the ellipse.

The mid point of the line segment joining the foci is called the centre of the ellipse. The line segment through the foci of the ellipse is called the major axis and the line segment through the centre and perpendicular to the major axis is called the minor axis. The end points of the major axis are called the vertices of the ellipse

We denote the length of the major axis by 2a, the length of the minor axis by 2b and the distance between the foci by 2c. Thus, the length of the semi major axis is a and semi-minor axis is b

Relationship between semi-major axis, semi-minor axis and the distance of the focus from the centre of the ellipse:

Sum of the distances of the point P to the foci is F1 P + F2P = F1O + OP + F2P

(Since, F1P = F1O + OP)

= c + a + a – c = 2a

a2 = b2 + c2

Special cases of an ellipse In the equation c2 = a2 – b2 obtained above, if we keep a fixed and vary c from 0 to a, the resulting ellipses will vary in shape.

Case (i) When c = 0, both foci merge together with the centre of the ellipse and a2 = b2, i.e.,  a = b, and so the ellipse becomes circle . Thus, circle is a special case of an ellipse.

Case (ii) When c = a, then b = 0. The ellipse reduces to the line segment F1F2 joining the two foci.

Eccentricity

Definition: The eccentricity of an ellipse is the ratio of the distances from the centre of the ellipse to one of the foci and to one of the vertices of the ellipse (eccentricity is  denoted by e) i.e., e= c/a

Standard equations of an ellipse The equation of an ellipse is simplest if the centre of the ellipse is at the origin and the foci are on the x-axis or y-axis.

Let F1 and F2 be the foci and O be the midpoint of the line segment F1F2. Let O be the origin and the line from O through F2 be the positive x-axis and that through F1as the negative x-axis.

Let, the line through O perpendicular to the x-axis be the y-axis. Let the coordinates of F1 be (– c, 0) and F2 be (c, 0) .

Let P(x, y) be any point on the ellipse such that the sum of the distances from P to the two foci be 2a

so given PF1 + PF2 = 2a.

Hence any point on the ellipse satisfies satisfies

the geometric condition and so P(x, y) lies on the ellipse.

Therefore, the ellipse lies between the lines x = – a and x = a and touches these lines.

Similarly, the ellipse lies between the lines y = – b and y = b and touches these lines.

Similarly, we can derive the equation of the ellipse.

These two equations are known as standard equations of the ellipses.

Properties:

1. Ellipse is symmetric with respect to both the coordinate axes since if (x, y) is a point on the ellipse, then (– x, y), (x, –y) and (– x, –y) are also points on the ellipse.

2. The foci always lie on the major axis. The major axis can be determined by finding the intercepts on the axes of symmetry. That is, major axis is along the x-axis if the coefficient of x2 has the larger denominator and it is along the y-axis if the coefficient of y2 has the larger denominator.

Latus rectum

Definition: Latus rectum of an ellipse is a line segment perpendicular to the major axis through any of the foci and whose end points lie on the ellipse.

Let the length of AF2 be l. Then the coordinates of A are (c, l ),

i.e., (ae, l )

Since A lies on the ellipse

we have

• l2 = b2 (1 – e2)   but e2=c2/a2 = 1- (b2/a2)

Therefore, l2 = b4/a2

• l = b2/a

Since the ellipse is symmetric with respect to y-axis (of course, it is symmetric w.r.t. both the coordinate axes), AF2 = F2B and so length of the latus rectum is 2l =2b2/a

Example:  Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the latus rectum of the ellipse

Solution Since denominator of x2/25 is larger than the denominator of y2/9 , the major

axis is along the x-axis.

a = 5 and b = 3. Also

c =√(a2 – b2)

=√(25 – 9)

=4

Therefore, the coordinates of the foci are (– 4,0) and (4,0), vertices are (– 5, 0) and

(5, 0). Length of the major axis is 10 units length of the minor axis 2b is 6 units and the

eccentricity is 4/5. and latus rectum is 2b2/a  = 18 /5

## 5. Standard equations, Properties and Application of a hyperbola

Standard equations , Properties and Application of a hyperbola

Hyperbola

Definition : A hyperbola is the set of all points in a plane, the difference of whose distances from two fixed points in the plane is a constant.

The term “difference” that is used in the definition means the distance to the farther point minus the distance to the closer point. The two fixed points are called the foci of the hyperbola. The mid-point of the line segment joining the foci is called the centre of the hyperbola. The line through the foci is called the transverse axis and the line through the centre and perpendicular to the transverse axis is called the conjugate axis. The points at which the hyperbola intersects the transverse axis are called the vertices of the hyperbola.

The distance between the two foci by 2c, the distance between two vertices (the length of the transverse axis) by 2a.

b= √(c2- a2)

2b is the length of the conjugate axis

we have

BF1 – BF2 = AF2 – AF1 (by the definition of the hyperbola)

BA +AF1– BF2 = AB + BF2– AF1

i.e., AF1 = BF2

So that, BF1 – BF2 = BA + AF1– BF2 = BA = 2a

Eccentricity

Definition : similarly, an ellipse, the ratio e = c / a is called the eccentricity of the

hyperbola. Since c ³a, the eccentricity is never less than one.

In terms of the eccentricity, the foci are at a distance of ae from the centre.

Standard equation of Hyperbola: The equation of a hyperbola is simplest if

the centre of the hyperbola is at the origin and the foci are on the x-axis or y-axis. The

two such possible orientations.

Let F1 and F2 be the foci and O be the mid-point of the line segment F1F2. Let O

be the origin and the line through O through F2 be the positive x-axis and

that through F1 as the negative x-axis.

Let P(x, y) be any point on the hyperbola such that the difference of the distances from P to the farther point minus the closer point be 2a.

So given, PF1 – PF2 = 2a

On squaring again and further simplifying, we get

(Since c2 a2 = b2)

Or

Note.: A hyperbola in which a = b is called an equilateral hyperbola.

Therefore, no portion of the curve lies between the

lines x = + a and x = – a, (i.e. no real intercept on the conjugate axis).

Similarly, we can derive the equation of the hyperbola.

These two equations are known as the standard equations of hyperbolas.

Properties:

1. Hyperbola is symmetric with respect to both the axes, since if (x, y) is a point on the hyperbola, then (– x, y), (x, – y) and (– x, – y) are also points on the hyperbola.
2. The foci are always on the transverse axis. It is the positive term whose denominator gives the transverse axis.

has transverse axis along x-axis of length 6.

has transverse axis along y-axis of length 10.

Latus rectum:

Definition Latus rectum of hyperbola is a line segment perpendicular to the transverse axis through any of the foci and whose end points lie on the hyperbola.

The length of the latus rectum in hyperbola is 2b2/a

Example Find the equation of the hyperbola where foci are (0, ±12) and the length of the latus rectum is 36.

Solution Since foci are (0, ±12), it follows that c = 12.

Length of the latus rectum = 2b2/a = 36

or b2 = 18a

Therefore c2 = a2 + b2; gives

144 = a2 + 18a

i.e., a2 + 18a – 144 = 0,

So a = – 24, 6.

Since a cannot be negative, we take a = 6 and so b2 = 108.

Therefore, the equation of the required hyperbola is

Or,

i.e., 3y2 – x2 = 108

Question :1 An ellipse passes through the foci of the hyperbola, 9x2 −4y2 =36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively. If the product of eccentricities of the two conics is 1/2. Find the equation of ellipse.

Solution:

Equation of hyperbola is 9x2 −4y2 =36 or x2/4 − y2/9 = 1

(Here a < b)

Focus = (0, ± be)

Eccentricity = e = √(1+4/9) = √13/3

So, Foci of hyperbola: (0, ±√13)

Standard equation of the ellipse, x2/a2 + y2/b2 = 1 …(i)

Eccentricity = e’ = √(1-a2/b2) …(ii)

ee’ = 1/2 (given)

Using eccentricity value of hyperbola, e’ = 1/2 x 3/√13 = 3/2√13

(ii) e’2 = (1-a2/b2)

9/52 = (1-a2/b2)

Find the value of b2 form (i) using focii 13/b2 = 1 => b2 = 13

9/52 = (1-a2/13)

9/4 = 13 – a2

a2 = 43/4

Now equation of ellipse is 4x2/43 + y2/13 = 1

Question 2: The locus of a point P(α, β) moving under the condition that the line y = αx + β is a tangent to the hyperbola x2/a2 – y2/b2 = 1 is

(a) an ellipse

(b) a circle

(c) a hyperbola

(d) a parabola