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 $P\left(3 cos \theta , 5 sin ⁡ \theta \right)$
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 $A\left(\frac{3}{c o s \theta } , 0\right)$ and $y$ -axis at $B\left(0 , \frac{5}{s i n \theta }\right)$
Let, $OA=OB=\lambda $
$cos \theta =\frac{3}{\lambda },sin⁡\theta =\frac{5}{\lambda }$
Now, $cos^{2}\theta +sin^{2}\theta =1$
$\frac{9}{\lambda ^{2}}+\frac{25}{\lambda ^{2}}=1\Rightarrow \lambda ^{2}=34$
$\therefore \lambda =\sqrt{34}$
Now, $AB=\sqrt{2}\lambda =2\sqrt{17}units$