Question

If any tangent to the ellipse $25x^2+9y^2=225$ meets the coordinate axes at $A$ and $B$ such that $OA=OB$ then, the length $AB$ is equal to (where, $O$ is the origin)

• A. $\sqrt{17}$ units
• B. $\sqrt{34}$ units
• C. $2\sqrt{17}$ units
• D. $2\sqrt{34}$ units

Answer 1

Answer: (C)

Given, equation of the ellipse is $\frac{x^2}{9}+\frac{y^2}{25}=1$
Let, a point on it be $P\left(3cos\theta ,5sin\theta \right)$
and equation of the tangent at $P$ is $\frac{x}{3}cos\theta +\frac{y}{5}sin\theta =1$
which meets the $x$ -axis at $A\left(\frac{3}{cos\theta },0\right)$ and $y$ -axis at $B\left(0,\frac{5}{sin\theta }\right)$
Let, $OA=OB=\lambda$
$cos\theta =\frac{3}{\lambda },sin\theta =\frac{5}{\lambda }$
Now, $co{s}^2\theta +si{n}^2\theta =1$
$\frac{9}{{\lambda }^2}+\frac{25}{{\lambda }^2}=1\Rightarrow {\lambda }^2=34$
$\therefore \lambda =\sqrt{34}$
Now, $AB=\sqrt{2}\lambda =2\sqrt{17}units$