From the point C(0, lambda),two tangents are | vedclass

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(B) The equation of the ellipse is $\frac{x^2}{4} + \frac{y^2}{2} = 1$. Here $a^2 = 4$ and $b^2 = 2$.Let the point of contact be $P(2\cos	heta, \sqrt

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(B) The equation of the ellipse is $\frac{x^2}{4} + \frac{y^2}{2} = 1$. Here $a^2 = 4$ and $b^2 = 2$.Let the point of contact be $P(2\cos	heta, \sqrt{2}\sin	heta)$.The equation of the tangent at $P$ is $\frac{x(2\cos	heta)}{4} + \frac{y(\sqrt{2}\sin	heta)}{2} = 1$,which simplifies to $\frac{x\cos	heta}{2} + \frac{y\sin	heta}{\sqrt{2}} = 1$.Since this tangent passes through $C(0, \lambda)$,we have $\frac{0\cdot\cos	heta}{2} + \frac{\lambda\sin	heta}{\sqrt{2}} = 1$,so $\sin	heta = \frac{\sqrt{2}}{\lambda}$.The tangent intersects the major axis $(y=0)$ at $A$. Setting $y=0$ in the tangent equation,we get $\frac{x\cos	heta}{2} = 1$,so $x_A = \frac{2}{\cos	heta}$.By symmetry,the other tangent intersects the major axis at $B(-x_A, 0)$.The base of $\Delta ABC$ is $AB = 2x_A = \frac{4}{\cos	heta}$ and the height is $\lambda$.Area $S = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes \frac{4}{\cos	heta} 	imes \lambda = \frac{2\lambda}{\cos	heta}$.Since $\sin	heta = \frac{\sqrt{2}}{\lambda}$,we have $\cos	heta = \sqrt{1 - \frac{2}{\lambda^2}} = \frac{\sqrt{\lambda^2 - 2}}{\lambda}$.Thus,$S = \frac{2\lambda^2}{\sqrt{\lambda^2 - 2}}$.To minimize $S$,let $u = \lambda^2$. Then $S^2 = \frac{4u^2}{u-2}$.Taking the derivative with respect to $u$: $\frac{d}{du}(\frac{4u^2}{u-2}) = 4 \frac{(u-2)(2u) - u^2(1)}{(u-2)^2} = 4 \frac{2u^2 - 4u - u^2}{(u-2)^2} = 4 \frac{u^2 - 4u}{(u-2)^2}$.Setting the derivative to $0$,we get $u(u-4) = 0$. Since $u = \lambda^2 > 2$,we have $u = 4$,so $\lambda^2 = 4$,which gives $\lambda = 2$.

From the point $C(0, \lambda)$,two tangents are drawn to the ellipse $x^2 + 2y^2 = 4$,cutting the major axis at $A$ and $B$. If the area of $\Delta ABC$ is minimum,then the value of $\lambda$ is-

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