Question

The ellipse having its foci $(0,\pm 1)$ and major axis of length $\sqrt{5}$ is

• A. $20x^2+4y^2=5$
• B. $36x^2+20y^2=45$
• C. $4c^2+20y^2=5$
• D. $20x^2+36y^2=45$

Answer 1

Answer: (A)

Since the foci of given ellipse lie on the $Y$-axis, it is vertical ellipse.
Let the required equation be
$\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$
Where $a^2>b^2$
Foci are $(0,\pm c)=(0,\pm 1)$
$\Rightarrow c=1$
$a=$ length of the semi-major axis
$=\frac{1}{2}\times \sqrt{5}=\frac{\sqrt{5}}{2}$
$\therefore a^2=\frac{5}{4}$
Now, $c^2=a^2-b^2$
$\Rightarrow 1={\left(\frac{\sqrt{5}}{2}\right)}^2-b^2$
$\Rightarrow b^2=\frac{5}{4}-1=\frac{1}{4}$
$\therefore$ The required equation is
$\frac{x^2}{1/4}+\frac{y^2}{5/4}=1$
$\Rightarrow 4x^2+\frac{4y^2}{5}=1$
$\Rightarrow 20x^2+4y^2=5$