The value of lambda,for which the line 2x - f | vedclass

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(A) The given conic is $x^2 + \frac{y^2}{4} = 1$,which is an ellipse with $a^2 = 1$ and $b^2 = 4$,so $a = 1$ and $b = 2$.The equation of the normal to

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$\frac{\sqrt{3}}{2}$

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(A) The given conic is $x^2 + \frac{y^2}{4} = 1$,which is an ellipse with $a^2 = 1$ and $b^2 = 4$,so $a = 1$ and $b = 2$.The equation of the normal to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at the point $(a \cos 	heta, b \sin 	heta)$ is given by $ax \sec 	heta - by \csc 	heta = a^2 - b^2$.Substituting $a = 1$ and $b = 2$,the normal equation becomes $x \sec 	heta - 2y \csc 	heta = 1 - 4 = -3$.Multiplying by $\cos 	heta$,we get $x - 2y \cot 	heta = -3 \cos 	heta$.The given line is $2x - \frac{8}{3}\lambda y = -3$. Rewriting this as $x - \frac{4}{3}\lambda y = -\frac{3}{2}$.Comparing the two equations,we have $\cot 	heta = \frac{2}{3}\lambda$ and $-3 \cos 	heta = -\frac{3}{2}$,which implies $\cos 	heta = \frac{1}{2}$.Since $\cos 	heta = \frac{1}{2}$,then $\sin 	heta = \frac{\sqrt{3}}{2}$ and $\cot 	heta = \frac{1}{\sqrt{3}}$.Equating $\frac{2}{3}\lambda = \frac{1}{\sqrt{3}}$ gives $\lambda = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}$.Wait,re-evaluating the comparison: The normal equation is $\frac{ax}{\cos 	heta} - \frac{by}{\sin 	heta} = a^2 - b^2$. For $a=1, b=2$,this is $\frac{x}{\cos 	heta} - \frac{2y}{\sin 	heta} = -3$.Comparing $\frac{x}{\cos 	heta} - \frac{2y}{\sin 	heta} = -3$ with $2x - \frac{8}{3}\lambda y = -3$,we have $\frac{1}{\cos 	heta} = \frac{2}{2} = 1 \implies \cos 	heta = 1$ (not possible) or scaling the equation.Scaling the normal equation by $2$: $\frac{2x}{\cos 	heta} - \frac{4y}{\sin 	heta} = -6$. This does not match. Correct approach: $\frac{x}{\cos 	heta} = \frac{2}{k}$ and $\frac{2y}{\sin 	heta} = \frac{8\lambda y}{3k}$ and $-3 = -3k$. Thus $k=1$. Then $\cos 	heta = 1/2$ and $\sin 	heta = 2 / (8\lambda/3) = 6/(8\lambda) = 3/(4\lambda)$.Since $\cos^2 	heta + \sin^2 	heta = 1$,$(1/2)^2 + (3/4\lambda)^2 = 1 \implies 9/(16\lambda^2) = 3/4 \implies 16\lambda^2 = 12 \implies \lambda^2 = 3/4 \implies \lambda = \sqrt{3}/2$.

The value of $\lambda$,for which the line $2x - \frac{8}{3}\lambda y = -3$ is a normal to the conic $x^2 + \frac{y^2}{4} = 1$ is

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