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

TS EAMCET 2020

Solution:

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 $
$\Rightarrow 4 x^{2}+\frac{4 y^{2}}{5}=1$
$\Rightarrow 20 x^{2}+4 y^{2}=5$