Question

The locus of poles of the tangents of $x^2+y^2=d^2$ w.r to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

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

Answer 1

The correct option is A $\frac{x^2}{a^4}+\frac{y^2}{b^4}=\frac{1}{d^2}$

The equation of the ellipse is
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Equation of the chord of contact of the tangents
drawn from the $P(x_1,y_1)$ to the ellipse is
$\frac{x{x}_1}{a^2}+\frac{y{y}_1}{b^2}=1$
Since the chord is at a distance d from the centre (0,0)
$d=\frac{\pm 1}{\sqrt{\frac{x_1^2}{a^4}+\frac{y_1^4}{b^2}}}$
$\frac{x_1^2}{a^4}+\frac{y_1^2}{b^4}=\frac{1}{d^2}$
Hence , Locus of the Point $P(x_1,y_1)$ is

$\frac{x^2}{a^4}+\frac{y^2}{b^4}=\frac{1}{d^2}$
Thus , Option A