## Example I

In this page, 'Example I' we are going to see how to get the equation of tangent to the given ellipse from a point (x₁, y₁)

#### Example I:

Find the equation of tangent to the ellipse x²/4 + y²/12 =1 at the point (-1,3).

Solution:

The equation of tangent at (x₁, y₁) is xx₁/a² + yy₁/b² = 1.

Here (x₁, y₁)  is (-1,3) and a =2 and b = √12.

The required equation of tangent is

x(-1)/4 + y(3)/12  = 1

-x/4    + y/4        = 1

Simplifying

-x﻿ + y       = 4

x - y + 4 = 0

The above equation is the required equation of the tangent to the given ellipse.﻿

#### Example 2:

Find the equation of the tangent to the ellipse x²+2y² =6 at (2,-1).

Solution:
The equation of tangent at (x₁, y₁) is xx₁/a² + yy₁/b² = 1.

The given equation is            x²+2y² =6

x²/6 + 2y²/6    =1

x²/6 +   y²/3    =1

Here (x₁, y₁) = (2, -1), a² =6 and b² = 3

The required equation of tangent is

x(2)/6  +  y(-1)/3  = 1

x/3    -    y/3      =  1

x-y           =  3

x-y-3     =  0

The above equation is the required equation of the tangent to the given ellipse.﻿

#### Example 3:

If the tangent at P of the ellipse x²/a² + y²/b² =1 meets the major axis at T and PN is the ordinate of , then prove that CN.CT = a² where C is the centre of the ellipse.

Solution:

Let P be the point on the ellipse whose coordinates are (x₁, y₁).

Equation of tangent at P(x₁, y₁) is xx₁/a² + yy₁/b² = 1

It meets the major axis which is y=0

Therefore the equation of the tangent becomes

xx₁/a²  =1

x          = a²/x₁

CT         =  a²/x

CN . CT    =  x₁ . a²/x₁

=   a².﻿

#### Example 4:

Find the equation of the tangents to the ellipse x²/16 + y²/9 =1 from (7,-3)

Solution:

A tangent to the given ellipse is of the form

y = mx + √(a²m²+b²)

Here a²=16 and b²=9

y  =  mx + √(16m²+9)

The tangent passes through (7,-3)

-3   = m(7) +  √(16m²+9)

Simplifying

-3-7m     =   √(16m²+9)

﻿

(-3-7m)²   =       16m² + 9

9+49m²+42m  =       16m² + 9

33m² +42m  =       0

3m(11m+14) =       0

m  =       0,  -14/11

The two tangents are

y = (0)x + √(16(0)² +9)  and

y = (-14/11)x+√(16(-14/11)²+9)

Simplifying

y = 3  and y = (-14/11)x+65/11

y = 3  and 11y  = -14x+65

y = 3 and  14x+11y = 65 are the required equations of tangents.

#### Practice problems:

1. Find the equation of tangent to the ellipse 4x²+9y²=36 at (2,2).
2. Find the equation of the tangent to the ellipse 3x²+2y²=5 at (-1,1).
3. Find the equation of the tangent to the ellipse x²/16 + y²/12 =1 at (2,-3).

Parents and teachers can guide the students to go through the examples discussed in page 'Example I' step by step. Students can solve the problems using the same method discussed above. If you are having any doubt you can contact us through mail, we will help you to clear your doubts.