# Math Help - Equation of an ellipse.

1. ## Equation of an ellipse.

find h in the ellipse $hy^2-4x^2+h=0$ if $a^2+b^2=10$
where a, b are the axes of the ellipse.

2. ## Re: Equation of an ellipse.

Originally Posted by razemsoft21
find h in the ellipse $hy^2-4x^2+h=0$ if $a^2+b^2=10$
where a, b are the axes of the ellipse.
There is no value of h for which $hy^2-4x^2+h=0$ is an ellipse. h > 0 defines a hyperbola, h < 0 is the null set, h = 0 is a line.

3. ## Re: Equation of an ellipse.

I agree with mr.fantastic, but let $h>0$ then you have a hyperbola. You know the equation of a hyperbola in general:
$\frac{x^2}{a^x}-\frac{y^2}{b^2}=1$

Now, for $h>0$:
Consider the given equation:
$hy^2-4x^2+h=0$
$\Leftrightarrow hy^2-4x^2=-h$
$\Leftrightarrow \frac{hy^2}{-h}-\left(\frac{4x^2}{-h}\right)=1$ (because h>0, we may divide the sides by h and also -h)
$\Leftrightarrow \frac{x^2}{\frac{h}{4}}-y^2=1$

Now we have the standard equation of the hyperbola and that means you can determine $a^2$ and $b^2$.

4. ## Re: Equation of an ellipse.

Originally Posted by Siron
I agree with mr.fantastic, but let $h>0$ then you have a hyperbola. You know the equation of a hyperbola in general:
$\frac{x^2}{a^x}-\frac{y^2}{b^2}=1$

Now, for $h>0$:
Consider the given equation:
$hy^2-4x^2+h=0$
$\Leftrightarrow hy^2-4x^2=-h$
$\Leftrightarrow \frac{hy^2}{-h}-\left(\frac{4x^2}{-h}\right)=1$ (because h>0, we may divide the sides by h and also -h)
$\Leftrightarrow \frac{x^2}{\frac{h}{4}}-y^2=1$

Now we have the standard equation of the hyperbola and that means you can determine $a^2$ and $b^2$.
I had assumed as much. And I suppose the OP giving Thanks is implicit confirmation of what the correct question was meant to be.

However, I was hoping that the OP would explicitly state the correction to his/her original question (which was the intention of my post).

5. ## Re: Equation of an ellipse.

If the "-" was simply 'misplaced' and the equation was $hy^2+ 4x^2- h= 0$, then we can rewrite it as
$4x^2+ hy^2= h$ and, dividing by h, $\frac{x^2}{\frac{h}{4}}+ y^2= 1$ 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 $a^2= \frac{h}{4}$ and $b^2= 1$.

From the first equation $h= 4a^2$ and from the second $b^2= 1$. $a^2= 10- b^2= 9$ so $h= 4(9)= 36$.