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If the normal to the ellipse $3x^2 + 4y^2 = 12$ 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
$\frac{{\sqrt {157} }}{2}$
$\frac{{5\sqrt 5 }}{2}$
$\frac{{\sqrt {221} }}{2}$
$\frac{{\sqrt {61} }}{2}$
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 $ Equation of noraml is $\frac{{{a^2}x}}{{{x_1}}} + \frac{{{b^2}y}}{{{y_1}}} = {a^2} – {b^2}$
$2\,x\,\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 $(4,4)$
$3\,x\cos \theta + 2\sqrt 3 \sin \theta y = 6$
$12\cos \theta + 8\sqrt 3 \sin \theta y = 6$
$\cos \theta = – \frac{1}{2},\sin \theta = \frac{{\sqrt 3 }}{2}\therefore {120^o}$
Hence point $P$ is $\left( {2\cos \theta {{120}^o},\sqrt 3 \sin {{120}^o}} \right)$
$P\left( { – 1,\frac{2}{2}} \right),Q\left( {4,4} \right)$
$PQ = \frac{{5\sqrt 5 }}{2}$
