question_answer

If the normal to the ellipse \[3{{x}^{2}}+4{{y}^{2}}=12\] at a point P on it is parallel to the line, \[2x+y=4\] and the tangent to the ellipse at P passes through Q(4, 4) then PQ is equal to:            [JEE Main Held on 12-4-2019 Morning]

A) \[\frac{\sqrt{221}}{2}\]                     

B) \[\frac{\sqrt{157}}{2}\]

C) \[\frac{\sqrt{61}}{2}\]                       

D) \[\frac{5\sqrt{5}}{2}\]

Correct Answer: D

Solution :

\[3{{x}^{2}}+4{{y}^{2}}=12\]           \[\frac{{{x}^{2}}}{4}+\frac{{{y}^{2}}}{3}=1\]           \[x=2\cos \theta ,y=\sqrt{3}\sin \theta \]           Let\[P(2\cos \theta ,\sqrt{3\sin \theta })\]             Equation of normal is\[\frac{{{a}^{2}}x}{{{x}_{1}}}-\frac{{{b}^{2}}y}{{{y}_{1}}}={{a}^{2}}-{{b}^{2}}\]             \[2x\sin \theta -\sqrt{3}\cos \theta y=\sin \theta \cos \theta \] Slope \[\frac{2}{\sqrt{3}}\tan \theta =-2\therefore \tan \theta =-\sqrt{3}\] Equation of tangent is it passes through (4, 4) \[3x\cos \theta +2\sqrt{3}\sin \theta y=6\] \[12\cos \theta +8\sqrt{3}\sin \theta =6\] \[\cos \theta =-\frac{1}{2},sin\theta =\frac{\sqrt{3}}{2}\therefore \theta ={{120}^{o}}\] Hence point P is \[\left( 2\cos 120{}^\circ ,\sqrt{3}\text{sin}120{}^\circ  \right)\] \[P\left( -1,\frac{3}{2} \right),Q(4,4)\] \[PQ=\frac{5\sqrt{5}}{2}\]