Question

If the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(b>a)$ and the parabola $y^2=4ax$ cut at right angles, then eccentricity of the ellipse is:

• A. $\frac{3}{5}$
• B. $\frac{2}{3}$
• C. $\frac{1}{\sqrt{2}}$
• D. $\frac{1}{2}$

Answer 1

Answer: (B), (C)

The correct options are
B $\frac{2}{3}$
C $\frac{1}{\sqrt{2}}$

Given, ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(b>a)$ and the parabola $y^2=4ax$ cut at right angles
That means tangents to the curves are perpendicular to each other.
Slope of tangent to ellipse at $(x_1,y_1)$ is $-\frac{b^2{x}_1}{a^2{y}_1}$
Slope of tangent to parabola at $(x_1,y_1)$ is $\frac{2a}{y_1}$
$\Rightarrow -\frac{b^2{x}_1}{a^2{y}_1}\frac{2a}{y_1}=-1$ $----1)$
$\Rightarrow (b^2-2a^2)x_1=0$
$\Rightarrow \frac{b^2}{a^2}=2$
$\Rightarrow \frac{a^2}{b^2}=\frac{1}{2}$
Hence, $e=\sqrt{1-\frac{1}{2}}=\frac{1}{\sqrt{2}}$