$p (θ), D (θ + \frac{π}{2})$ are two points on the Ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ Then the locus of point of intersection of the two tangents at P and D to the ellipse is

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Solution

The correct option is C
Equation of tangents at P and D are
$\frac{x}{a} c o s θ + \frac{y}{b} s i n θ = 1 - (1) - \frac{x}{a} s i n θ + \frac{y}{b} c o s θ = 1 - (2)$
Squaring and adding 1 and 2, we get.
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 2$

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Similar questions

P (θ)

Q (θ + \frac{π}{2})

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1.

P (θ), D (θ + \frac{π}{2})

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

P D

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

x^{2} - y^{2} = 0

\frac{x^{2}}{a^{2} + b^{2}} + \frac{y^{2}}{b^{2}} = 1

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{a^{2} + b^{2}} = 1

\frac{x^{2}}{a^{2} + b^{2}} + \frac{y^{2}}{b^{2}} = 1, \frac{x^{2}}{a^{2}} + \frac{y^{2}}{a^{2} + b^{2}} = 1

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