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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(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-

Similar Questions

If $\frac{\pi}{3}$ and $\theta$ are the eccentric angles of the ends of a focal chord of the ellipse $\frac{x^2}{16}+\frac{y^2}{12}=1$,then $\tan \theta=$

Let $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ $(b < a)$ be an ellipse with major axis $AB$ and minor axis $CD$. Let $F_1$ and $F_2$ be its two foci,with $A, F_1, F_2, B$ in that order on the segment $AB$. Suppose $\angle F_1CB = 90^{\circ}$. The eccentricity of the ellipse is:

The locus of the foot of the perpendicular drawn from the centre of the ellipse $x^2 + 3y^2 = 6$ to any tangent to it is:

If tangents are drawn to the ellipse $x^2 + 2y^2 = 2$ at all points on the ellipse other than its four vertices,then the midpoints of the tangents intercepted between the coordinate axes lie on the curve:

The value of $k$,if $(1, 2)$ and $(k, -1)$ are conjugate points with respect to the ellipse $2x^2 + 3y^2 = 6$,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module