Question

Find the equation of ellipse with foci at (+5,0) and (-5,0) and

x = 36/5 as one of its directrix.

Answer 1

Required equation of Ellipse :

x^2/36 + y^2/11 = 1 .


For solution see attachment.

Answer 2

The required equation is x²/36 + y²/11 = 1

Let the eq. of the  ellipse be x²/a²+y²/b² = 1

Let the be eccentricity be = e

Coordinates of focii = (±5,0)

Equations of one of the  directrix = x= 36/5

So, ae=5 and ae=36/5

= ae × a/e =5 × 36/5

= a² = 36

= a = 6

Now, b² = a²(1−e²)

= b² = a²−(ae)² = 36−25 = 11

b = √11

Substituting a=6 and b = 11 in  x²/a²+y²/b²=1 we will get -

x²/36 + y²/11 = 1