Question

What is the equation of the ellipse with `foci (0, 1+-\sqrt(8); vertices (0, -2), (0,4)`?

Answer 1

Equation of ellipse is `9x^2+8(y-1)^2=72`

Explanation:

As focii are aligned with `y`-axis, we have a vertical ellipse, whose equation is of type `(x-h)^2/a^2+(y-k)^2/b^2=1`, where major axis along `y`-axis is `2b` and minor axis is `2a`.

It is observed that midpoint of focii aswell as vertices both is `(0,1)` and as for all of them abscissa is same, we have a vertical ellipse and major axis is `4-(-2)=6` units.

Hence equation of ellipse is of the form.

`(x-0)^2/a^2+(y-1)^2/(6/2)^2=1`

Further distance between focii is `2sqrt8` which is equal to major axis multiplied by eccentricity. Hence eccentricity `e=(2sqrt8)/6=(2sqrt2)/3`

and hence `a=3xx(2sqrt2)/3=2sqrt2`

and equation of ellipse is

`(x-0)^2/(2sqrt2)^2+(y-1)^2/(6/2)^2=1`

or `x^2/8+(y-1)^2/9=1`

or `9x^2+8(y-1)^2=72`

graph{(9x^2+8(y-1)^2-72)(x^2+(y-1+sqrt8)^2-0.02)(x^2+(y-1-sqrt8)^2-0.02)=0 [-10.5, 9.5, -4, 6]}