find h in the ellipse $\displaystyle hy^2-4x^2+h=0$ if $\displaystyle a^2+b^2=10$
where a, b are the axes of the ellipse.
I agree with mr.fantastic, but let $\displaystyle h>0$ then you have a hyperbola. You know the equation of a hyperbola in general:
$\displaystyle \frac{x^2}{a^x}-\frac{y^2}{b^2}=1$
Now, for $\displaystyle h>0$:
Consider the given equation:
$\displaystyle hy^2-4x^2+h=0$
$\displaystyle \Leftrightarrow hy^2-4x^2=-h$
$\displaystyle \Leftrightarrow \frac{hy^2}{-h}-\left(\frac{4x^2}{-h}\right)=1$ (because h>0, we may divide the sides by h and also -h)
$\displaystyle \Leftrightarrow \frac{x^2}{\frac{h}{4}}-y^2=1$
Now we have the standard equation of the hyperbola and that means you can determine $\displaystyle a^2$ and $\displaystyle b^2$.
If the "-" was simply 'misplaced' and the equation was $\displaystyle hy^2+ 4x^2- h= 0$, then we can rewrite it as
$\displaystyle 4x^2+ hy^2= h$ and, dividing by h, $\displaystyle \frac{x^2}{\frac{h}{4}}+ y^2= 1$ which, for h> 0, is an ellipse. Assuming that "a" is the length of the semi-axis along the x-axis and "b" the length of the sem-axis along the y-axis, we have $\displaystyle a^2= \frac{h}{4}$ and $\displaystyle b^2= 1$.
From the first equation $\displaystyle h= 4a^2$ and from the second $\displaystyle b^2= 1$. $\displaystyle a^2= 10- b^2= 9$ so $\displaystyle h= 4(9)= 36$.