Question

A line perpendicular to the $X$ - axis cuts the circle $x^{2}+y^{2}=9$ at $A$ and the ellipse $4 x^{2}+9 y^{2}=36$ at $B$ such that $A$ and $B$ lie in the same quadrant. If $\theta$ is the greatest acute angle between the tangents drawn to the curves at $A$ and $B$, then $\tan \theta=$

• A. $\frac{1}{12}$
• B. $\frac{1}{2 \sqrt{6}}$
• C. $\frac{5}{24}$
• D. $\frac{5}{4 \sqrt{6}}$

Answer 1

Answer: (B)

Let the equation of line perpendicular to $X$ -axis cut the circle $x^{2}+y^{2}=9$ at $A$ is $(3 \cos \alpha, 3 \sin \alpha)$
Equation of tangent at $A$ is $x \cos \alpha+y \sin \alpha=3$
Slope $=-\cot \alpha$
Similarly, cut the ellipse $4 x^{2}+9 y^{2}=36$ at $B$ is $(3 \cos \alpha, 2 \sin \alpha)$
Equation of tangent at $B$
$2 \cos \alpha+3 \sin \alpha =6 $
Slope$=-\frac{2}{3} \cot \alpha$
Angle between tangents is $\theta$, then
$\tan \,\theta=\frac{\left(-\frac{2}{3}+1\right) \cot \alpha}{1+\frac{2}{3} \cot ^{2} \alpha}=\frac{\cot \alpha}{3+2 \cot ^{2} \alpha}$
Let $\tan \theta=z $
$\Rightarrow z=\frac{\cot \alpha}{3+2 \cot ^{2} \alpha}$
$\because \frac{d z}{d \alpha}=-\left[\frac{\left(3+2 \cot ^{2} \alpha\right)\left(\text{cosec}^{2} \alpha\right)-\cot \alpha\left(4 \cot \alpha \text{cosec}^{2} \alpha\right)}{\left(3+2 \cot ^{2} \alpha\right)}\right]$
For maxima or minima $\frac{d z}{d \alpha}=0$
$\therefore 3 \text{cosec}^{2} \alpha+2 \cot ^{2} \alpha \text{cosec}^{2} \alpha-4 \cot ^{2} \alpha \text{cosec}^{2} \alpha=0$
$\Rightarrow \cos ^{2} \alpha=\frac{3}{2} $
$\Rightarrow \cot \alpha=\sqrt{\frac{3}{2}}$
$\because \tan \theta=\frac{\sqrt{\frac{3}{2}}}{3+2 \times \frac{3}{2}}=\sqrt{\frac{3}{2}} \times \frac{1}{6}=\frac{1}{2 \sqrt{6}}$
$\therefore $ Greatest acute angle when, $\tan \theta=\frac{1}{2 \sqrt{6}}$.