Question

If the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ meets the line $\frac{x}{7}+\frac{y}{2 \sqrt{6}}=1$ on the $x$-axis and the line $\frac{x}{7}-\frac{y}{2 \sqrt{6}}=1$ on the $y$-axis, then the eccentricity of the ellipse is

• A. $\frac{5}{7}$
• B. $\frac{2 \sqrt{6}}{7}$
• C. $\frac{3}{7}$
• D. $\frac{2 \sqrt{5}}{7}$

Answer 1

Answer: (A)

Ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ passes through the point $(7,0)$ and $(0,-2 \sqrt{6})$
Now $\frac{49}{ a ^2}+0=1 \Rightarrow a ^2=49$
and $0+\frac{24}{ b ^2}=1 \Rightarrow b ^2=24$
Now $a > b \Rightarrow b ^2= a ^2\left(1- e ^2\right)$
$\Rightarrow 24=49\left(1- e ^2\right) \Rightarrow e ^2=\frac{25}{49} \Rightarrow e =\frac{5}{7}$