Question

Let $A\left(\alpha \right)$ and $B\left(\beta \right)$ be the extremities of a chord of an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. If the slope of $AB$ is equal to the slope of the tangent at a point $C\left(\theta \right)$ on the ellipse, then the value of $\theta$ is

• A. $\frac{\alpha -\beta }{2}$
• B. $\frac{\alpha -\beta }{2}-\pi$
• C. $\frac{\alpha +\beta }{2}$
• D. $\frac{\alpha +\beta }{2}+\pi$

Answer 1

Answer: (C), (D)

$\frac{\alpha +\beta }{2}+\pi$

Given ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ $A(a\cos \alpha ,b\sin \alpha ),B(a\cos \beta ,b\sin \beta )$ and $C(a\cos \theta ,b\sin \theta )$
Equation of tangent at $C$ is $T=0$
$\Rightarrow \frac{x\cos \theta }{a}+\frac{y\sin \theta }{b}=1$
Slope of $AB=$ Slope of tangent at $C$
$\Rightarrow \frac{b(\sin \beta -\sin \alpha )}{a(\cos \beta -\cos \alpha )}=\frac{-b\cos \theta }{a\sin \theta }$
$\Rightarrow \frac{2\cos \left(\frac{\beta +\alpha }{2}\right)\sin \left(\frac{\beta -\alpha }{2}\right)}{2\sin \left(\frac{\alpha +\beta }{2}\right)\sin \left(\frac{\alpha -\beta }{2}\right)}=\frac{-\cos \theta }{\sin \theta }$
$\Rightarrow \frac{-\cos \left(\frac{\alpha +\beta }{2}\right)}{\sin \left(\frac{\alpha +\beta }{2}\right)}=\frac{-\cos \theta }{\sin \theta }$
$\Rightarrow \tan \left(\frac{\alpha +\beta }{2}\right)=\tan \theta$
$\Rightarrow \theta =\frac{\alpha +\beta }{2}+n\pi (n\in I)$.
Hence, $\theta =\frac{\alpha +\beta }{2}$ or $\frac{\alpha +\beta }{2}+\pi$