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 $2 h$, 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 $=2 h$
$\therefore 2 a e=2 h \Rightarrow a e=h...$(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
$\frac{x^{2}}{k^{2}}+\frac{y^{2}}{k^{2}-h^{2}}=1$