Ellipse Eqn, given focal radius of 8, a^2/b^2 = 3/2

[cyberspace]

Just to make sure, the general equation for an ellipse is:

(x-h)^2 / a^2 + (y-k)^2 / b^2 = 1

And my teacher, who made up this problem, assigns this for homework:

Given the focal radius= 8, and a^2/b^2= 3/2, find the equation of the ellipse.

Note: a^2/b^2= 3/2 is expressed as a ratio.

Thanks for the help!

 

[soroban]

Hello, cyberspace!

\(\displaystyle \text{Given the focal radius 8, and }\frac{a^2}{b^2} = \frac{3}{2},\:\text{ find the equation of the ellipse.}\)

I will assume that the ellipse is centered at the origin.

\(\displaystyle \text{We are given: }\:\frac{a^2}{b^2}\:=\:\frac{3}{2}\quad\Rightarrow\quad b^2 \:=\:\frac{2}{3}a^2\)


\(\displaystyle \text{The "focal equation" is: }\:a^2 \;=\;b^2+c^2\)

\(\displaystyle \text{Substitute: }\;a^2\;=\;\frac{2}{3}a^2 + 8^2\quad\Rightarrow\quad \frac{1}{3}a^2\;=\;64\quad\Rightarrow\quad\boxed{ a^2\:=\:192}\)

\(\displaystyle \text{Then: }\:b^2 \:=\:\frac{2}{3}(192) \quad\Rightarrow\quad\boxed{ b^2 \:=\:128}\quad \hdots\quad There!\)


. . \(\displaystyle \text{The equation is: }\;\frac{x^2}{192} + \frac{y^2}{128} \;=\;1\)

 

[cyberspace]

Thank you so much for the help! :mrgreen: