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 $a>b\Rightarrow b^2=a^2\left(1-e^2\right)$
$\Rightarrow 24=49\left(1-e^2\right)\Rightarrow e^2=\frac{25}{49}\Rightarrow e=\frac{5}{7}$