Question

The the focal distance of an end of the minor axis of any ellipse (reffered to its 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)

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

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 $k$.

$\therefore a+e\left(0\right)=k\Rightarrow a-k=0$ ...(2)

From (1) and (2)

$b^2=a^2(1-e^2)=k^2-h^2$

$\therefore$ The equation of the ellipse is

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