Find the equation of the ellipse referred to its centre whose minor axis is equal to the distance between the foci and whose latus is 10?

Question

$\begin{array}{1 1}(a)\;\large\frac{x^2}{100}+\frac{y^2}{50}\normalsize=1\\(b)\;\large\frac{x^2}{10}+\frac{y^2}{5}\normalsize=1\\(c)\;\large\frac{x^2}{200}+\frac{y^2}{100}\normalsize=1\\(d)\;\text{None of these}\end{array}$

Answer 1

Let the equation of the ellipse is $\large\frac{x^2}{a^2}+\frac{y^2}{b^2}$$=1\qquad a > b$

Then the foci are $S(ae,0)$ and $S'(-ae,0)$

Length of minor axes =2b

Length of lactus rectum =$\large\frac{2b^2}{a}$

According to the question

$BB'=SS'$

$2b=2ae$

$b=ae$

$\large\frac{2b^2}{a}$$=10$

$b^2=5a$

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

$a^2e^2=a^2(1-e^2)$

$e=\large\frac{1}{\sqrt 2}$

$b=\large\frac{a}{\sqrt 2}$

$b^2=\large\frac{a^2}{2}$

$5a=\large\frac{a^2}{2}$

$a=10$

$b^2=5\times 10=50$

Hence equation of ellipse

$\large\frac{x^2}{100}+\frac{y^2}{50}$$=1$

Hence (a) is the correct answer.