Question

How do you find the equation of the ellipse which satisfies the following conditions: the foci are at (+-2,0) and the ellipse passes through (2,-3)?

Answer 1

Equation of ellipse is `x^2/16+y^2/12=1`

Explanation:

As the foci are `(+2,0)` and `(-2,0)`, the major axis is aligned with `x`-axis and centre of the ellipse is `(0,0)`. The equation of such an ellipse is of the type

`x^2/a^2+y^2/b^2=1`, where `2a` is major axis and `2b` is minor axis and `b^2=a^2(1-e^2)=a^2-a^2e^2`. where `e` is the eccentricity of the ellipse and foci are `(ae,0)` and `(-ae,0)`. Here `ae=2` and `a>2`.

Then the equation becomes `x^2/a^2+y^2/(a^2(1-e^2))=1`

As it passes through `(2,-3)`, we have

`4/a^2+9/(a^2-4)=1`

or `4(a^2-4)+9a^2=a^2(a^2-4)` and if `a^2=k` it becomes

`4k-16+9k=k^2-4k`

or `k^2-17k+16=0`

or `(k-1)(k-16)=0`

i.e. `k=a^2=1` or `k=a^2=16`

As `a>2`, we have `a=4` and `b=sqrt(4^2-2^2)sqrt12`

and equation of ellipse is `x^2/16+y^2/12=1`

graph{x^2/16+y^2/12=1 [-10, 10, -5, 5]}