If the normal to the ellipse 3x^2 + 4y^2 = 12 | vedclass

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Question

$3{x^2} + 4{y^2} = 12$

$\frac{{{x^2}}}{4} + \frac{{{y^2}}}{3} = 1$

$x = 2\,\cos 	heta ,y = \sqrt 3 \sin 	heta $ Equation of noraml is $\f

Vedclass · Accepted Answer

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

Vedclass · Answer

$3{x^2} + 4{y^2} = 12$

$\frac{{{x^2}}}{4} + \frac{{{y^2}}}{3} = 1$

$x = 2\,\cos 	heta ,y = \sqrt 3 \sin 	heta $ Equation of noraml is $\frac{{{a^2}x}}{{{x_1}}} + \frac{{{b^2}y}}{{{y_1}}} = {a^2} - {b^2}$

$2\,x\,\sin 	heta  - \sqrt 3 \cos 	heta y = \sin 	heta \cos 	heta $

Slope $\frac{2}{{\sqrt 3 }}	an 	heta  =  - 2\,\,\,\,\,$            $	herefore 	an 	heta  =  - \sqrt 3 $

equation of tangent is it passes through $(4,4)$

$3\,x\cos 	heta  + 2\sqrt 3 \sin 	heta y = 6$

$12\cos 	heta  + 8\sqrt 3 \sin 	heta y = 6$

$\cos 	heta  =  - \frac{1}{2},\sin 	heta  = \frac{{\sqrt 3 }}{2}	herefore {120^o}$

Hence point $P$ is $\left( {2\cos 	heta {{120}^o},\sqrt 3 \sin {{120}^o}} ight)$

$P\left( { - 1,\frac{2}{2}} ight),Q\left( {4,4} ight)$

$PQ = \frac{{5\sqrt 5 }}{2}$

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

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