Question

The point of intersection of tangents to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at the points where the line $x\cos \alpha +y\sin \alpha =p$ meets it, is

• A. $(\frac{a^2\sin \alpha }{p},\frac{b^2\cos \alpha }{p})$
• B. $(\frac{a^2\cos \alpha }{p^2},\frac{b^2\sin \alpha }{p^2})$
• C. $(\frac{a^2\cos \alpha }{p},\frac{b^2\sin \alpha }{p})$
• D. None of these

Answer 1

The correct option is A $(\frac{a^2\sin \alpha }{p},\frac{b^2\cos \alpha }{p})$

$\begin{array}{c}\frac{hx}{a^2}+\frac{ky}{b^2}=1\\ \frac{\cos \alpha }{p}x+\frac{siny}{p}=1\\ Oncomparingaboveequations\\ \frac{h}{a^2}=\frac{\cos \alpha }{p}\\ \frac{k}{b^2}=\frac{\sin \alpha }{p}\\ h=\frac{a^2\cos \alpha }{p}k=\frac{b^2\sin \alpha }{p}\\ {\left(\frac{hp}{a^2}\right)}^2+{\left(\frac{kp}{b^2}\right)}^2-1\\ \left(h,k\right)=\left(\frac{a^2\cos \alpha }{p},\frac{b^2\sin \alpha }{p}\right)\\ OptionCiscorrectanswer.\end{array}$