If the ellipse x^2a^2+y^2b^2=1 meets the line x7+ y2√6=1 on the x-axis and the line x7- y2…

Question

If the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ meets the line $\dfrac x7+\dfrac y{2\sqrt6}=1$ on the $x$-axis and the line $\dfrac x7-\dfrac y{2\sqrt6}=1$ on the $y$-axis, then the eccentricity of the ellipse is

Accepted Answer

First line meets the $x$-axis at $(7,0)$, so $a=7$. Second line meets the $y$-axis at $(0,-2\sqrt6)$, so $b=2\sqrt6$, $b^2=24$.

$$e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{24}{49}}=\sqrt{\frac{25}{49}}=\frac57.$$

Solutions · 1