Question

An ellipse having co-ordinate axes as its axes having lengths $2a$ and $2b$ units respectively, Where $a$ and $b$ are middle terms of a series $a_1,a_2,a_3\cdots a_{10},$ $a_i{a}_{11-i}=5\sqrt{3}\forall i\in N,i<11.$ If $B,F,F^{\prime }$ are one end of minor axis and foci of the ellipse respectively,such that triangle $FB{F}^{\prime }$ is an equilateral triangle, then equation of ellipse is

• A. $\frac{x^2}{10}+\frac{2y^2}{15}=1$
• B. $\frac{x^2}{15}+\frac{y^2}{5}=1$
• C. $\frac{x^2}{10}+\frac{y^2}{15}=1$
• D. $\frac{2x^2}{15}+\frac{y^2}{10}=1$

Answer 1

The correct option is A $\frac{x^2}{10}+\frac{2y^2}{15}=1$

Let ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
As, $a$ and $b$ are the middle term of $a_1,a_2,...a_{10}$
$\therefore a=a_5,b=a_6$
We have
$a_i{a}_{11-i}=5\sqrt{3}$
$\Rightarrow a_5\cdot a_6=5\sqrt{3}$
$\therefore ab=5\sqrt{3}\cdots \left(1\right)$
Now, $FB{F}^{\prime }$ is an equilateral triangle
$\therefore F{F}^{\prime }=FB$
$\Rightarrow 2ae=\sqrt{a^2{e}^2+b^2}$
$\Rightarrow 2ae=a\Rightarrow e=\frac{1}{2}$
$\because a^2-b^2=a^2{e}^2$
$\Rightarrow b^2=\frac{3a^2}{4}\cdots \left(2\right)$
From equation $\left(1\right)$ and $\left(2\right),$ we get
$a^2=10,b^2=\frac{15}{2}$
$\therefore$ The equation of the ellipse is
$\frac{x^2}{10}+\frac{2y^2}{15}=1$