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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}

 
Answers (1)
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V Vakul

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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