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

Answer: (C)

Let the equation of the ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.
Let $e$ be the eccentricity of the ellipse.
Since distance between foci $=2h$
$\therefore 2ae=2h\Rightarrow ae=h...$(1)
Focal distance of one end of minor axis say $(0,b)$ is $k$.
$\therefore a+e(0)=k\Rightarrow a=k...$(2)
From (1) and (2)
$b^2=a^2\left(1-e^2\right)=k^2-h^2$
$\therefore$ The equation of the ellipse is
$\frac{x^2}{k^2}+\frac{y^2}{k^2-h^2}=1$