Question

An ellipse is rotated through a right angle in its own plane about its centre, which is fixed. If at a point tangents are drawn before and after rotation then locus of point of intersection of such tangents is

• A. $(x^2+y^2)(x^2+y^2-a^2-b^2)=2(a^2-b^2)xy$
• B. $(x^2+y^2)(a^2+b^2)=2(a^2-b^2)xy$
• C. $(x^2+y^2)(a^2-b^2)=2(a^2+b^2)xy$
• D. $(x^2+y^2)(x^2+y^2-a^2-b^2)=2(a^2+b^2)xy$

Answer 1

The correct option is A $(x^2+y^2)(x^2+y^2-a^2-b^2)=2(a^2-b^2)xy$

Let the coordinates of point $P$ on the ellipse be $(a\cos \theta ,b\sin \theta )$. After the rotation be a right angle in the anticlockwise direction, the very point becomes $Q(-b\sin \theta ,a\cos \theta ).$ And the new equation of the ellipse is $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$
Equation of tangent at $P$ is $\frac{x\cos \theta }{a}+\frac{y\sin \theta }{b}=1\cdots \left(1\right)$
Equation of tangent at $Q$ is :
$\Rightarrow \frac{-x\sin \theta }{b}+\frac{y\cos \theta }{a}=1\cdots \left(2\right)$
Let the point be intersection be $(h,k)$
Solving $\left(1\right)$ and $\left(2\right)$, we get
$h=\frac{\frac{\cos \theta }{a}-\frac{\sin \theta }{b}}{\frac{{\cos }^2\theta }{a^2}+\frac{{\sin }^2\theta }{b^2}}$ and
$k=\frac{\frac{\cos \theta }{a}+\frac{\sin \theta }{b}}{\frac{{\cos }^2\theta }{a^2}+\frac{{\sin }^2\theta }{b^2}}$
Squaring and adding we get:
$h^2+k^2=\frac{1}{\frac{{\cos }^2\theta }{a^2}+\frac{{\sin }^2\theta }{b^2}}$
$\therefore \frac{h}{h^2+k^2}=\frac{\cos \theta }{a}-\frac{\sin \theta }{b}$ and
$\frac{k}{h^2+k^2}=\frac{\cos \theta }{a}+\frac{\sin \theta }{b}$
Eliminating $\theta$, we get
$(h^2+k^2)(h^2+k^2-a^2-b^2)=2(a^2-b^2)hk$
Hence locus will be
$(x^2+y^2)(x^2+y^2-a^2-b^2)=2(a^2-b^2)xy$