Question

If the focal distance of an end of the minor axis of any ellipse (referred to its axes as the axes of $x$ and $y$ respectively) is $k$ and the distance between the foci is $2h$, then its equation is:

• A. $\frac{x^2}{k^2}+\frac{y^2}{k^2+h^2}=1$
• B. $\frac{x^2}{k^2}+\frac{y^2}{h^2-k^2}=1$
• C. $\frac{x^2}{k^2}+\frac{y^2}{k^2-h^2}=1$
• D. $\frac{x^2}{k^2}+\frac{y^2}{h^2}=1$

Answer 1

The correct option is A $\frac{x^2}{k^2}+\frac{y^2}{k^2+h^2}=1$

Distance between foci$=2h$

Then, $2ae=2h$

$ae=h$

and $b=k$

$b^2=k^2$

We know that, $b^2=a^2(1-e^2)$

$b^2=a^2-a^2{e}^2$

Substituting the values, we get

$k^2=a^2-h^2$

$a^2=k^2{+h}^2$

Hence equation of ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

$\frac{x^2}{k^2+h^2}+\frac{y^2}{b^2}=1$

OR

$\frac{x^2}{k^2}+\frac{y^2}{k^2+h^2}=1$