Q

# How to solve this problem- - Co-ordinate geometry - JEE Main-10

If the normal to the ellipse $3x^{2}+4y^{2}=12$ at a point P on its parallel to the line, $2x+y=4$ and the tangent to the ellipse at P passes through  $Q\left ( 4,4 \right )$ then PQ is equal to :

• Option 1)

$\frac{\sqrt{157}}{2}$

• Option 2)

$\frac{\sqrt{221}}{2}$

• Option 3)

$\frac{\sqrt{61}}{2}$

• Option 4)

$\frac{5\sqrt{5}}{2}$

Views

$3x^{2}+4y^{2}=12$

$\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$

So,  $x=2\cos \theta \: \: \: \: \: \: \: \: \: \: \: \: y=\sqrt{3}\sin \theta$

Let               $P\left ( 2\cos \theta ,\sqrt{3}\sin \theta \right )$

Equation of normal is $\frac{a^{2}x}{x_{1}}-\frac{b^{2}y}{y_{1}}=a^{2}-b^{2}$

$\frac{4x}{2\cos \theta }-\frac{3y}{\sqrt{3}\sin \theta }=4-3$

$2x\sin \theta -\sqrt{3}y\cos \theta =\cos \theta \sin \theta$

$Slope=\frac{2}{\sqrt{3}}\tan \theta =-2$

$\tan \theta =-\sqrt{3}$

Equation of tangent is it passes through $(4,4)$

$3x\cos \theta +2\sqrt{3}\sin \theta \: y=6$

$12\cos \theta +8\sqrt{3}\sin \theta =6$

$\frac{1}{2}\cos \theta +\frac{8\sqrt{3}}{6}\sin \theta =1$

$\cos \theta =-\frac{1}{2},\: \: \: \sin \theta =\frac{\sqrt{3}}{2}\: \: \: \: \therefore \theta =120^{\circ}$

Hence point is $\left ( 2\cos 120,\sqrt{3}\sin 120 \right )$

$P\left ( -1,\frac{3}{2} \right ),Q\left ( 4,4 \right )$

$PQ=\sqrt{\left ( -1-4 \right )^{2}+\left ( \frac{3}{2}-4 \right )^{2}}=\sqrt{25+\frac{25}{4}}$

$=\frac{5\sqrt{5}}{2}$

Option 1)

$\frac{\sqrt{157}}{2}$

Option 2)

$\frac{\sqrt{221}}{2}$

Option 3)

$\frac{\sqrt{61}}{2}$

Option 4)

$\frac{5\sqrt{5}}{2}$

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