find the ellipse 2

Question

write the eq. of the ellipse w/ center at the origin, its foci on the x axis, which passes through the points (-3, 2 sq. root of 3) and (4, 4 sq. root of 5/ 3)

Answer 1

(2) The equation can be written: x²/a²+y²/b²=1.

Plug in the given points and we get two equations:

9/a²+12/b²=1, 16a²+80/(9b²)=1.

From these: a²=9b²/(b²-12) and a²=144b²/(9b²-80).

Therefore 9b²/(b²-12)=144b²/(9b²-80), 1/(b²-12)=16/(9b²-80),

Cross-multiplying: 9b²-80=16b²-192, 7b²=112, b²=16, b=4.

So a²=9b²/(b²-12)=144/4=36, a=6.

The equation of the ellipse is x²/36+y²/16=1.

This can also be written: 16x²+36y²=576.