Question

The locus of point of intersection of tangent to an ellipse at two points, sum of whose eccentric angle is constant is

• A. Parabola
• B. Circle
• C. Ellipse
• D. Straight line

Answer 1

The correct option is D Straight line

The equation of tangents at two points having eccentric angle ${\theta }_1$ and ${\theta }_2$ are

$\frac{x}{a}\cos {\theta }_1+\frac{y}{b}\sin {\theta }_1=1....\left(i\right)$

and $\frac{x}{a}\cos {\theta }_2+\frac{y}{b}\sin {\theta }_2....\left(ii\right)$

The point of intersection of Eqs. (i) and (ii) is

$(\frac{a\cos \left(\frac{{\theta }_1+{\theta }_2}{2}\right)}{\cos \left(\frac{{\theta }_1-{\theta }_2}{2}\right)},\frac{b\sin \left(\frac{{\theta }_1+{\theta }_2}{2}\right)}{\cos \left(\frac{{\theta }_1-{\theta }_2}{2}\right)})$

It is given that ${\theta }_1+{\theta }_2=k=constant$.

Therefore, if $(x_1,y_1)$ is the point of intersection of Eqs. (i) and (ii), then

$x_1=\frac{a\cos \frac{k}{2}}{\cos \left(\frac{{\theta }_1-{\theta }_2}{2}\right)}$

and

$y_1=\frac{b\sin \frac{k}{2}}{\cos \left(\frac{{\theta }_1-{\theta }_2}{2}\right)}$

$\Rightarrow \frac{x_1}{y_1}=\frac{a}{b}\cot \left(\frac{k}{2}\right)$

$y_1=\left(\frac{b}{a}\cot \left(\frac{k}{2}\right)\right)x_1$

$\Rightarrow (x_1,y_1)$ lies on the straight line $y=\left(\frac{b}{a}\cot \left(\frac{k}{2}\right)\right)x$