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

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(B) The equation of the ellipse is $3x^2 + 4y^2 = 12$,which can be written as $\frac{x^2}{4} + \frac{y^2}{3} = 1$. Here $a^2 = 4$ and $b^2 = 3$.Let th

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

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(B) The equation of the ellipse is $3x^2 + 4y^2 = 12$,which can be written as $\frac{x^2}{4} + \frac{y^2}{3} = 1$. Here $a^2 = 4$ and $b^2 = 3$.Let the point $P$ be $(2\cos 	heta, \sqrt{3}\sin 	heta)$.The equation of the normal at $P$ is $\frac{a^2x}{x_1} - \frac{b^2y}{y_1} = a^2 - b^2$,which simplifies to $\frac{4x}{2\cos 	heta} - \frac{3y}{\sqrt{3}\sin 	heta} = 4 - 3$,or $2x\sec 	heta - \sqrt{3}y\csc 	heta = 1$.The slope of this normal is $\frac{2\sec 	heta}{\sqrt{3}\csc 	heta} = \frac{2}{\sqrt{3}}	an 	heta$.Since the normal is parallel to $2x + y = 4$,its slope is $-2$. Thus,$\frac{2}{\sqrt{3}}	an 	heta = -2$,which gives $	an 	heta = -\sqrt{3}$.For $	an 	heta = -\sqrt{3}$,we have $\cos 	heta = \pm \frac{1}{2}$ and $\sin 	heta = \mp \frac{\sqrt{3}}{2}$.The equation of the tangent at $P$ is $\frac{x\cos 	heta}{2} + \frac{y\sin 	heta}{\sqrt{3}} = 1$,or $x\cos 	heta + \frac{2y\sin 	heta}{\sqrt{3}} = 2$.Since the tangent passes through $Q(4, 4)$,we have $4\cos 	heta + \frac{8\sin 	heta}{\sqrt{3}} = 2$,or $2\sqrt{3}\cos 	heta + 4\sin 	heta = \sqrt{3}$.Substituting $\sin 	heta = -\sqrt{3}\cos 	heta$,we get $2\sqrt{3}\cos 	heta - 4\sqrt{3}\cos 	heta = \sqrt{3}$,so $-2\sqrt{3}\cos 	heta = \sqrt{3}$,which means $\cos 	heta = -\frac{1}{2}$ and $\sin 	heta = \frac{\sqrt{3}}{2}$.Thus,$P = (2(-1/2), \sqrt{3}(\sqrt{3}/2)) = (-1, 3/2)$.$PQ = \sqrt{(4 - (-1))^2 + (4 - 3/2)^2} = \sqrt{5^2 + (5/2)^2} = \sqrt{25 + 25/4} = \sqrt{125/4} = \frac{5\sqrt{5}}{2}$.

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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