Mathematics - Ellipse Question with Solution | TestHub

Given,

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ meets the line $\frac{x}{7}+\frac{y}{2\sqrt{6}}=1$ on the $x$-axis, so at $x-$axis $y=0$ putting in both equation and comparing we get, $a=7$ 

And $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ meets the line $\frac{x}{7}-\frac{y}{2\sqrt{6}}=1$ on the $y$-axis where $x=0$ so putting in both equation and comparing we get, $b=2\sqrt{6}$

Therefore, $e^2=1-\frac{b^2}{a^2}=1-\frac{24}{49}$

$\Rightarrow e=\sqrt{\frac{25}{49}}=\frac{5}{7}$