Question

An ellipse passing through $(4\surd 2,2\surd 6)$ has foci at $(-4,0)$ and $(4,0)$. Then, its eccentricity is

• A. $\sqrt{2}$
• B. $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$
• C. $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.$
• D. $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{3}$}\right.$

Answer 1

The correct option is B

$\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

Step 1. Given data:

Foci are$(-4,0)$ and $(-4,0)$

In an ellipse, the sum of the distance of any point on the ellipse from the focii equals $2a$.

and the distance between the focii equals $2ae$.

$\Rightarrow 2ae=8$

$\therefore ae=4$

Step 2. As we know that:

The ellipse is$x^2{a}^2+y^2{b}^2=1$

$(4\surd 2,2\surd 6)$ lies on it

$\therefore \frac{32}{a^2}+\frac{24}{b^2}=1$

where,

$$\begin{gathered} b^2=a^2(1–e^2) \\ \Rightarrow b^2=a^2–a^2{e}^2 \\ \Rightarrow b^2=a^2–a^2{e}^2 \\ \Rightarrow b^2=a^2–16 \\ \Rightarrow \frac{32}{a^2}+\frac{24}{(a^2–16)}=1 \\ \Rightarrow \frac{24}{(a^2–16)}=1–\left(\frac{32}{a^2}\right) \\ \Rightarrow \frac{24}{(a^2–16)}=\frac{[a^2–32a^2]}{a^2} \\ \Rightarrow 24a^2=a^4–48a^2+512 \\ \Rightarrow a^4–72a^2+512=0 \\ \Rightarrow (a^2–64)(a^2–8)=0 \\ \Rightarrow a^2=64 \\ \Rightarrow a=8 \\ \Rightarrow 8e=4 \\ \therefore e=\frac{1}{2} \end{gathered}$$

Hence, the correct option is (B).