Question

If in an ellipse, whose centre is at origin, twice the length of the minor axis =the distance between the foci and its latus rectum = $10$, then the equation of the ellipse in the standard form is:

• A. $\frac{x^2}{{\left(10\right)}^2}+\frac{y^2}{{\left(5\sqrt{5}\right)}^2}=1$
• B. $\frac{x^2}{{\left(5\sqrt{2}\right)}^2}+\frac{y^2}{{\left(10\right)}^2}=1$
• C. $\frac{x^2}{25}+\frac{y^2}{{\left(5/\sqrt{2}\right)}^2}=1$
• D. none of these

Answer 1

The correct option is A $\frac{x^2}{{\left(10\right)}^2}+\frac{y^2}{{\left(5\sqrt{5}\right)}^2}=1$

In an ellipse,

Let major axis length = $2a$

and minor axis length = $2b$

Now,

Given, $2ae=2b$

$ae=b$ $⟶\left(1\right)$

and $\frac{{2b}^2}{a}=10⟶\left(2\right)$

we know that

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

$b^2=a^2-a^2{e}^2$

$b^2+a^2{e}^2=a^2$

from (1), $b^2=a^2{e}^2$

$\therefore {2b}^2=a^2$

putting this in eqn(2)

$\frac{a^2}{a}=10,a=10$

putting value of a in eqn(2)

${2b}^2=100$

$b=\sqrt[5]{52}$

Therefore equation of ellipse : $\frac{x^2}{{10}^2}+\frac{y^2}{{\left(\sqrt[5]{52}\right)}^2}=1$