Question

The locus of point of intersection of tangents drawn at $\alpha ,\beta$ on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ such that $\alpha -\beta =\frac{\pi }{3}$ is:

• A. $3\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}\right)=4$
• B. $4\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}\right)=3$
• C. $\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}\right)=2$
• D. $\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}\right)=4$

Answer 1

The correct option is A $3\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}\right)=4$

Point of intersection of the two tangents drwan at $\alpha ,\beta$ are:

$x=\frac{a\cos \left(\frac{\alpha +\beta }{2}\right)}{\cos \left(\frac{\alpha -\beta }{2}\right)}$ and $y=\frac{b\sin \left(\frac{\alpha +\beta }{2}\right)}{\cos \left(\frac{\alpha -\beta }{2}\right)}$

Given: $\alpha -\beta =\frac{\pi }{3}$

So, $h=\frac{a\cos \left(\frac{\alpha +\beta }{2}\right)}{\cos \left(\frac{\pi }{6}\right)}$ and $k=\frac{b\sin \left(\frac{\alpha +\beta }{2}\right)}{\cos \left(\frac{\pi }{6}\right)}$

$\Rightarrow \frac{\sqrt{3}h}{2a}=\cos \left(\frac{\alpha +\beta }{2}\right)$ and $\frac{\sqrt{3}k}{2a}=\sin \left(\frac{\alpha +\beta }{2}\right)$

On squaring and adding both the equations we get:

$\frac{{3h}^2}{{4a}^2}+\frac{{3k}^2}{{4b}^2}=1$

$\therefore$ The locus is: $3(\frac{x^2}{a^2}+\frac{y^2}{b^2})=4$