Question

The locus of the point of intersection of tangents to an ellipse at two points whose eccentric angles differ by a constant $\alpha$ is

• A. $x^2+y^2=a^2{b}^2\tan \alpha$
• B. $\frac{x^2}{a^2}+\frac{y^2}{b^2}=\sec \alpha$
• C. $b^2{x}^2+a^2{y}^2=a^2{b}^2{\sec }^2\frac{\alpha }{2}$
• D. None of these

Answer 1

The correct option is C $b^2{x}^2+a^2{y}^2=a^2{b}^2{\sec }^2\frac{\alpha }{2}$

For two points in general on an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with eccentric angle $\beta$ and $\gamma$, the coordinates of point of intersection of tangents is $(a\frac{cos\frac{\beta +\gamma }{2}}{cos\frac{\beta -\gamma }{2}},b\frac{sin\frac{\beta +\gamma }{2}}{cos\frac{\beta -\gamma }{2}})$
Here given that $\beta -\gamma =\alpha$
So, $x=a\frac{cos\frac{\beta +\gamma }{2}}{cos\frac{\beta -\gamma }{2}}$
$\Rightarrow xcos\frac{\alpha }{2}=acos\frac{\beta +\gamma }{2}$
Similarly, $ycos\frac{\alpha }{2}=bsin\frac{\beta +\gamma }{2}$

Now using the identity $si{n}^2\frac{\beta +\gamma }{2}+co{s}^2\frac{\beta +\gamma }{2}=1$, we get

$\frac{x^2{\cos }^2\frac{\alpha }{2}}{a^2}+\frac{y^2{\cos }^2\frac{\alpha }{2}}{b^2}=1$

$\Rightarrow b^2{x}^2+a^2{y}^2=a^2{b}^{2}se{c}^2\frac{\alpha }{2}$