Question

If the sum of the eccentric angles of two points of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $2\alpha$ (constant), then the locus of point of intersection of the two tangents at these points is

• A. $ay=bx\tan \alpha$
• B. $ax=by\tan \alpha$
• C. $ay=bx\cot \alpha$
• D. $ax=by\cot \alpha$

Answer 1

The correct option is A $ay=bx\tan \alpha$

Apply the concepts of tangent in terms of point of contact and parametric representation
Let $\varphi ,{\varphi }_1$ be eccentric angle of these two points
from question $\varphi +{\varphi }_1=2\alpha$
Equation of first tangent can be $\frac{xa\cos \varphi }{a^2}+\frac{yb\sin \varphi }{b^2}-1=0$
Equation of second tangent can be $\frac{xa\cos {\varphi }_1}{a^2}+\frac{yb\sin {\varphi }_1}{b^2}-1=0$
$⟹$
Equation of first tangent can be $\frac{x\cos \varphi }{a}+\frac{y\sin \varphi }{b}-1=0$
Equation of second tangent can be $\frac{x\cos {\varphi }_1}{a}+\frac{y\sin {\varphi }_1}{b}-1=0$
point of intersection will be
$\frac{x}{-\frac{\sin \varphi }{b}+\frac{\sin {\varphi }_1}{b}}=\frac{y}{-\frac{\cos {\varphi }_1}{a}+\frac{\cos \varphi }{a}}=\frac{1}{\frac{\cos \varphi \sin {\varphi }_1}{ab}-\frac{\cos {\varphi }_1\sin \varphi }{ab}}=\frac{ab}{\sin ({\varphi }_1-\varphi )}$
$x=\frac{a(\sin {\varphi }_1-{\sin }_{\varphi })}{\sin ({\varphi }_1-\varphi )}=\frac{2a\cos \left(\frac{\varphi +{\varphi }_1}{2}\right)\sin \left(\frac{{\varphi }_1-\varphi }{2}\right)}{\sin ({\varphi }_1-\varphi )}$
$y=\frac{b(\cos \varphi -\cos {\varphi }_1)}{\sin ({\varphi }_1-\varphi )}=\frac{2b\sin \left(\frac{\varphi +{\varphi }_1}{2}\right)\sin \left(\frac{{\varphi }_1-\varphi }{2}\right)}{\sin ({\varphi }_1-\varphi )}$
$⟹\frac{x}{y}=\frac{a}{b}\cot \left(\frac{\varphi +{\varphi }_1}{2}\right)=\frac{a\cot \alpha }{b}$
$bx\tan \alpha =ya$