Question

The point of intersection of tangents at two points on the ellipse $\dfrac{{{x^2}}}{{{a^2}}}\, + \dfrac{{{y^2}}}{{{b^2}}} = 1$, whose eccentric angles differ by right angle lies on the ellipse, is
A) $\dfrac{{{x^2}}}{{{a^2}}}\, + \dfrac{{{y^2}}}{{{b^2}}} = 4$
B) $\dfrac{{{x^2}}}{{{a^2}}}\, + \dfrac{{{y^2}}}{{{b^2}}} = 3$
C) $\dfrac{{{x^2}}}{{{a^2}}}\, + \dfrac{{{y^2}}}{{{b^2}}} = 2$
D) None of these

Answer 1

An ellipse is a plane curve surrounding two focal points. For all those points, the sum of the two distances to the focal points is a constant. For two points in general, on ellipse $\dfrac{{{x^2}}}{{{a^2}}}\, + \dfrac{{{y^2}}}{{{b^2}}} = 1$ with eccentric angles $\beta$and $\gamma$. The coordinates of point of intersection is $\left( {a\dfrac{{\cos \dfrac{{\beta + \gamma }}{2}}}{{\cos \dfrac{{\beta - \gamma }}{2}}},b\dfrac{{\sin \dfrac{{\beta + \gamma }}{2}}}{{\cos \dfrac{{\beta - \gamma }}{2}}}} \right)$.

Complete step-by-step solution:
Let the first point on ellipse be A whose coordinates are given by $\left( {a\,\cos \theta ,b\sin \theta } \right)$
Now the second on ellipse, whose parametric angle differs by $\dfrac{\pi }{2}$ from A be B. so the coordinates of B are $\left( {a\,\cos \theta \left( {\theta + \dfrac{\pi }{2}} \right),b\sin \left( {\theta + \dfrac{\pi }{2}} \right)} \right) \equiv \left( { - a\,\sin \theta ,b\,\cos \theta } \right)$
Now equation of tangent at point A is given by, $\dfrac{{x\,\cos \theta }}{a} + \dfrac{{y\,\sin \theta }}{b} = 1$- equation1
The equation of tangent at point B is given by, $\dfrac{{ - x\,\sin \theta }}{a} + \dfrac{{y\,\cos \theta }}{b} = 1$-equation2
To find the intersection points of locus and square of equation 1 and equation 2, we get:
${\left( {\dfrac{{x\,\cos \theta }}{a} + \dfrac{{y\,\sin \theta }}{b}} \right)^2} + {\left( {\dfrac{{ - x\sin \theta }}{a} + \dfrac{{y\,\cos \theta }}{b}} \right)^2} = 2$
Further solving the equation, we get:
$\dfrac{{{x^2}{{\cos }^2}\theta }}{{{a^2}}} + \dfrac{{{y^2}{{\sin }^2}\theta }}{{{b^2}}} + \dfrac{{2xy\sin \theta \,\cos \theta }}{{ab}} + \dfrac{{{x^2}{{\sin }^2}\theta }}{{{a^2}}} + \dfrac{{{y^2}{{\cos }^2}\theta }}{{{b^2}}} - \dfrac{{2xy\,\sin \theta \cos \theta }}{{ab}} = 2$
Third and last expressions get cancelled. Cancelling the terms, we get:
$\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 2$
So, Option C is the right answer.

Note: In this case we already know the answer, so it is evident if we square and add the two equations, we will get the answer. However, if we wouldn’t have known the answer, or this method is not striking your mind easily, then there is another way to solve this.