Question

The eccentricity of the ellipse which meets the straight line $\frac{x}{7}+\frac{y}{2}=1$ on the axis of $x$ and the straight line $\frac{x}{3}-\frac{y}{5}=1$ on the axis of $y$ and whose axes lie along the axes of coordinates is

• A. $\frac{\sqrt{6}}{7}$
• B. $\frac{4 \sqrt{6}}{7}$
• C. $\frac{2 \sqrt{6}}{5}$
• D. $\frac{2 \sqrt{6}}{7}$

Answer 1

Answer: (D)

Line $\frac{x}{7}+\frac{y}{2}=1$ meet $x$-axis at, $y=0$
$\Rightarrow x=7$
line $\frac{x}{3}-\frac{y}{5}=1$ meet $y$-axis at $ x=0 \Rightarrow y=-5$
$\therefore$ ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ passes through $(7,0)$ and $(0,-5)$
hence $\frac{49}{a^2}+0=1 \Rightarrow a^2=49$
$0+\frac{25}{b^2}=1 \Rightarrow b^2=25$
$\therefore \frac{x^2}{49}+\frac{y^2}{25}=1 \Rightarrow e=\sqrt{1-\frac{25}{49}}$