Question

An Ellipse is rotated through a right angle in its own plane about its center. Prove that the locus of POI of a tangent to the ellipse in its Original Position with the Tangent at the same point on the Curve in its new Position is: $(x^2+y^2)(x^2+y^2-a^2-b^2)=2(a^2-b^2)xy$

Answer 1

The locus of the point of intersection is given by the equation: $(x^2+y^2)(x^2+y^2-a^2-b^2) = 2(a^2-b^2)xy$

Answer 2

1. Parametric Representation and Tangent: The original ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ has a tangent $T_1$ at $P(a \cos \theta, b \sin \theta)$ given by $\frac{x \cos \theta}{a} + \frac{y \sin \theta}{b} = 1$.

2. Rotated Ellipse and Tangent: After a $90^\circ$ rotation, the ellipse becomes $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$. The tangent $T_2$ to this rotated ellipse at the corresponding point $P_{rot}(-b \sin \theta, a \cos \theta)$ is $-\frac{x \sin \theta}{b} + \frac{y \cos \theta}{a} = 1$.

3. Point of Intersection and Locus: Solving for $\cos \theta$ and $\sin \theta$ from the system of equations for $T_1$ and $T_2$, and using $\cos^2 \theta + \sin^2 \theta = 1$, yields the locus equation: $\left(\frac{a(x+y)}{x^2+y^2}\right)^2 + \left(\frac{b(y-x)}{x^2+y^2}\right)^2 = 1$ This simplifies to the required locus.