Question

The locus of chords of contact of perpendicular tangents to the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}$= 1 touch another fixed ellipse is-

• A. $\frac{x^{2}}{a^{2}}$+ $\frac{y^{2}}{b^{2}}$ = $\frac{1}{(2a^{2} + b^{2})}$
• B. $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = \frac{2}{(a^{2}–b^{2})}$
• C. $\frac{x^{2}}{a^{4}} + \frac{y^{2}}{b^{4}} = \frac{1}{(a^{2} + b^{2})}$
• D. $\frac{x^{2}}{a^{2}}–\frac{y^{2}}{b^{2}} = \frac{2}{(3a^{2}–b^{2})}$

Answer 1

Answer: (C)

$\frac{x^{2}}{a^{4}} + \frac{y^{2}}{b^{4}} = \frac{1}{(a^{2} + b^{2})}$

Answer 2

We know that locus of the point of intersection of perpendicular tangents to the given ellipse is x² + y² = a² + b² .

Any point on this circle can be taken as

P ŗ$\left( \sqrt{a^{2} + b^{2}}\cos\theta,\sqrt{a^{2} + b^{2}}\sin\theta \right)$The equation of the chord of contact of tangents from P is

$\frac{x}{a^{2}}\sqrt{a^{2} + b^{2}}\cos{}\theta + \frac{y}{b^{2}}\sqrt{a^{2} + b^{2}}$sin q = 1.

Let this line be a tangent to the fixed ellipse $\frac{x^{2}}{A^{2}} + \frac{y^{2}}{B^{2}}$ = 1.

Ž $\frac{x}{A}$cos q + $\frac{y}{B ⥂}$sin q = 1,

Where A = $\frac{a^{2}}{\sqrt{a^{2} + b^{2}}}$, B = $\frac{b^{2}}{\sqrt{a^{2} + b^{2}}}$.

$\frac{x^{2}}{a^{4}} + \frac{y^{2}}{b^{4}} = \frac{1}{(a^{2} + b^{2})}$.

\ an ellipse.

Hence (3) is the correct answer.