 Comment
Share
Q)

# The equation of the ellipse having foci $(0,1),(0,-1)$ and minor axis of length $\sqrt 5$ is ____________

$\begin{array}{1 1}\large\frac{4x^2}{5}+\frac{4y^2}{1}\normalsize =1\\\large\frac{4x^2}{1}+\frac{4y^2}{5}\normalsize =1\\\large\frac{5x^2}{4}+\frac{y^2}{4}\normalsize =1\\\large\frac{x^2}{4}+\frac{5y^2}{4}\normalsize =1\end{array}$ • General equation of an ellipse whose axis is along the minor axis $\large\frac{x^2}{a^2}+\frac{y^2}{b^2}$$=1 where b > a • a^2=b^2(1-e^2) Answer : \large\frac{4x^2}{1}+\frac{4y^2}{5}\normalsize =1 Given foci is (0,\pm 1) Clearly the ellipse lies along the minor axis. Length of the minor axis is \sqrt 5 (ie) 2b=\sqrt 5 \therefore b=\large\frac{\sqrt 5}{2} Given foci is be=1 \Rightarrow \large\frac{\sqrt 5}{2}$$\times e=1$
$\therefore e=\large\frac{2}{\sqrt 5}$
Hence $a^2=b^2(1-e^2)$
$a^2=\large\frac{5}{4}$$(1-\large\frac{4}{5}) a^2=\large\frac{1}{4} and b^2=\large\frac{5}{4} Hence equation of the ellipse is \large\frac{x^2}{\Large\frac{1}{4}}+\frac{y^2}{\Large\frac{5}{4}}$$=1$
$\Rightarrow \large\frac{4x^2}{1}+\frac{4y^2}{5}$$=1$ is the required equation.