Question

The locus of the points of the intersection of tangents to ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ which make an angle $\theta$ is

• A. $x^2+y^2=4\tan \theta (a^2+b^2)$
• B. $(x^2+y^2)\tan \theta =b^2{x}^2+a^2{y}^2$
• C. $(x^2+y^2-a^2-b^2){\tan }^2\theta =(b^2{x}^2+a^2{y}^2-a^2{b}^2)$
• D. None of these

Answer 1

Answer: (C)

$(x^2+y^2-a^2-b^2){\tan }^2\theta =(b^2{x}^2+a^2{y}^2-a^2{b}^2)$

equation of pair of tangent from a point is given by $S{S}_1=T^2$
$(\frac{x^2}{a^2}+\frac{y^2}{b^2})(\frac{x_1^2}{a^2}+\frac{{y_1}^2}{b^2})={(\frac{x{x}_1}{a^2}+\frac{y{y}_1}{b^2})}^2$
Angle between pair of straight lines is given by $\tan \theta =\left|\frac{2\sqrt{h^2-ab}}{a+b}\right|$
Therefore, locus of $P$ is $(x^2+y^2-a^2-b^2){\tan }^2\theta =(b^2{x}^2+a^2{y}^2-a^2{b}^2)$