Question

If the line $x+2y+4=0$ cutting the ellipse in points whose eccentric angles are ${30}^{\circ }$ and ${60}^{\circ }$ subtends a right angle at the origin, then its equation is

• A. $\frac{x^2}{4}+\frac{y^2}{16}=1$
• B. $\frac{x^2}{16}+\frac{y^2}{4}=1$
• C. $\frac{x^2}{8}+\frac{y^2}{4}=1$
• D. $\frac{x^2}{4}+\frac{y^2}{8}=1$

Answer 1

The correct option is B $\frac{x^2}{16}+\frac{y^2}{4}=1$

Let $P(a\cos {30}^{\circ },b\sin {30}^{\circ })$ and $Q$ be $(a\cos {60}^{\circ },b\sin {60}^{\circ })$
Its slope $=\frac{b(\sin {60}^{\circ }-\sin {30}^{\circ })}{a(\cos {60}^{\circ }-\cos {30}^{\circ })}=-\frac{b}{a}$
But slope of $x+2y+4=0$ is $-\frac{1}{2}$ $\therefore -\frac{b}{a}=-\frac{1}{2}$ or $a=2b$
The chord subtends an angle of ${90}^{\circ }$ at the origin.
Making equation of the ellipse homogeneous with the equation of line, we have $\frac{x^2}{a^2}+\frac{y^2}{b^2}={\left(\frac{x+2y}{-4}\right)}^2$ or $x^2(\frac{1}{a^2}-\frac{1}{16})-\frac{xy}{4}+y^2(\frac{1}{b^2}-\frac{1}{4})=0$
Since the line subtends an angle of ${90}^{\circ }$ at origin $\therefore A+B=0$ or $\frac{1}{a^2}-\frac{1}{16}+\frac{1}{b^2}-\frac{1}{4}=0$ or $\frac{1}{4b^2}-\frac{1}{16}+\frac{1}{b^2}-\frac{1}{4}=0$ $\because a=2b,by\left(1\right)$ $\therefore \frac{5}{4b^2}=\frac{5}{16}$ $\therefore b^2=4\therefore a^2=4b^2=16$
Hence the equation is $\frac{x^2}{16}+\frac{y^2}{4}=1$