Question

If the line $x+2y+4=0$ cutting the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ in points whose eccentric angles are ${30}^{\circ }$ and ${60}^{\circ }$ subtends a right angle at the origin then its equation is

• A. $\frac{x^2}{8}+\frac{y^2}{4}=1$
• B. $\frac{x^2}{16}+\frac{y^2}{4}=1$
• C. $\frac{x^2}{4}+\frac{y^2}{16}=1$
• D. none of these

Answer 1

The correct option is B

$\frac{x^2}{16}+\frac{y^2}{4}=1$

slope of line $=\frac{b}{a}\left(\frac{\sin {60}^{\circ }-\sin {30}^{\circ }}{\cos {60}^{\circ }-\cos {30}^{\circ }}\right)=-\frac{b}{a}=-\frac{1}{2}$
Homogenizing the ellipse with $x+2y-4=0$
$\frac{x^2}{4b^2}+\frac{y^2}{b^2}={\left(\frac{x+2y}{-4}\right)}^2\Rightarrow \frac{1}{b^2}(\frac{x^2}{4}+y^2)=\frac{x^2+4y^2+4xy}{16}$
Coefficient of $x^2$+coefficient of $y^2=0$
$\frac{1}{4b^2}-\frac{1}{16}+\frac{1}{b^2}-\frac{4}{16}=0\Rightarrow b^2=4,b=2\Rightarrow a=4\text{Ellipse is}\frac{x^2}{16}+\frac{y^2}{4}=1$