Question

Two perpendicular tangents drawn to the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ intersect on the curve.

• A. $x^2+y^2=41$
• B. $x^2+y^2=9$
• C. $x^2-y^2=41$
• D. $x^2-y^2=9$

Answer 1

The correct option is A $x^2+y^2=41$

Locus of point from where perpendicular tangents can be drawn to $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $x^2+y^2=a^2+b^2$ (Director circle)

Hence, two perpendicular tangents of $\frac{x^2}{25}+\frac{y^2}{16}=1$ intersect on $x^2+y^2=25+16$

$\Rightarrow x^2+y^2=41$.