Question

A line perpendicular to the $X$ - axis cuts the circle $x^2+y^2=9$ at $A$ and the ellipse $4x^2+9y^2=36$ 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 $4x^2+9y^2=36$ 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 $\tan \theta =z$
$\Rightarrow z=\frac{\cot \alpha }{3+2{\cot }^2\alpha }$
$\because \frac{dz}{d\alpha }=-\left[\frac{\left(3+2{\cot }^2\alpha \right)\left({\text{cosec}}^2\alpha \right)-\cot \alpha \left(4\cot \alpha {\text{cosec}}^2\alpha \right)}{\left(3+2{\cot }^2\alpha \right)}\right]$
For maxima or minima $\frac{dz}{d\alpha }=0$
$\therefore 3{\text{cosec}}^2\alpha +2{\cot }^2\alpha {\text{cosec}}^2\alpha -4{\cot }^2\alpha {\text{cosec}}^2\alpha =0$
$\Rightarrow {\cos }^2\alpha =\frac{3}{2}$
$\Rightarrow \cot \alpha =\sqrt{\frac{3}{2}}$
$\because \tan \theta =\frac{\sqrt{\frac{3}{2}}}{3+2\times \frac{3}{2}}=\sqrt{\frac{3}{2}}\times \frac{1}{6}=\frac{1}{2\sqrt{6}}$
$\therefore$ Greatest acute angle when, $\tan \theta =\frac{1}{2\sqrt{6}}$.