## Solution 1

In this page 'solution 1' we are going to see step by step methods of finding the equation of ellipse for the given foci, eccentricity and directrix.

Practice Problem 1:

Find the equation of ellipse whose focus is (1,2), directrix is 2x-3y+6=0 and eccentricity is 2/3.﻿

Solution for practice problem 1:

Focus             =    (1,2)

Directrix         =    2x-3y+6=0

Eccentricity     =    2/3

(x₁-1)² + (y₁-1)² =  (4/9)[(2x₁-3y₁+6)/√(4+9)]²

=  (4/(9*13))(2x₁-3y₁+6)²

9*13[x₁²-2x₁+1+y₁²-4y₁+4] = 4(4x₁²+9y₁²+36-12x₁y₁-36y₁+24x₁)

117x₁²-234x₁+117+117y₁²-468y₁+468 =

16x₁²+36y₁²+144-48x₁y₁-144y₁+96x₁

117x₁²-16x₁²-234x₁-96x₁+117y₁²-36y₁²-468y₁+144y₁+48x₁y₁+

117+468-144 = 0

101x₁²+81y₁²-330x₁-324y₁+48x₁y₁+441   = 0

The locus of (x₁,y₁), i.e., The equation of ellipse is

101x²+81y²-330x-324y+48xy+441   = 0                         ﻿

Practice Problem 2:

Find the equation of the ellipse whose foci are (4,0) and (-4,0) and e =1/3.

Solution for practice problem 2:

This problem is little different from the previous problem. In the previous problem one focus and directrix is given. But here two foci are given. So we follow little different method to solve this problem.

Foci  = (4,0) and (-4,0)

eccentricity = 1/3

ae   =  4

e   =   1/3

∴    a   =   4x3  = 12

b²  =   a²(1-e²)

=   12²(1-⅓²)

=   144(1-(1/9))

=   128

The standard equation of ellipse is

x²/a²   +   y²/b²   =1

So the required equation is

x²/144  +  y²/128 = 1﻿

Parents and teachers can guide the students to follow the solutions in this page 'Solution 1', step by step. Students can practice the problems in the methods discussed above and try to solve the problems given below on their own. If you have any doubt you can contact us through mail, we will help you to clear your doubts.

Problems for practice:

1. Find the equation of ellipse whose focus is (0,0), directrix 3x+4y-1=0 and eccentricity is 5/6.
2. Find the equation of the ellipse whose foci are (3,0) and (-3,0) and the eccentricity is 3/8,﻿