Let a tangent be drawn to the ellipse frac{x^ | vedclass

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

Question

(C) The equation of the tangent to the ellipse $\frac{x^{2}}{27}+y^{2}=1$ at the point $(3 \sqrt{3} \cos 	heta, \sin 	heta)$ is given by $\frac{x \c

Vedclass · Accepted Answer

$\frac{\pi}{6}$

Vedclass · Answer

(C) The equation of the tangent to the ellipse $\frac{x^{2}}{27}+y^{2}=1$ at the point $(3 \sqrt{3} \cos 	heta, \sin 	heta)$ is given by $\frac{x \cos 	heta}{3 \sqrt{3}}+\frac{y \sin 	heta}{1}=1$.The $x$-intercept is $OA = 3 \sqrt{3} \sec 	heta$ and the $y$-intercept is $OB = \operatorname{cosec} 	heta$.Let the sum of the intercepts be $f(	heta) = 3 \sqrt{3} \sec 	heta + \operatorname{cosec} 	heta$.To find the minimum,we differentiate $f(	heta)$ with respect to $	heta$:$f^{\prime}(	heta) = 3 \sqrt{3} \sec 	heta 	an 	heta - \operatorname{cosec} 	heta \cot 	heta$.Setting $f^{\prime}(	heta) = 0$:$3 \sqrt{3} \frac{\sin 	heta}{\cos^{2} 	heta} = \frac{\cos 	heta}{\sin^{2} 	heta}$$	an^{3} 	heta = \frac{1}{3 \sqrt{3}} = \left(\frac{1}{\sqrt{3}}ight)^{3}$$	an 	heta = \frac{1}{\sqrt{3}} \Rightarrow 	heta = \frac{\pi}{6}$.Since $f^{\prime}(	heta)  0$ for $	heta > \frac{\pi}{6}$,the function $f(	heta)$ attains its minimum value at $	heta = \frac{\pi}{6}$.

Let a tangent be drawn to the ellipse $\frac{x^{2}}{27}+y^{2}=1$ at $(3 \sqrt{3} \cos \theta, \sin \theta)$ where $\theta \in\left(0, \frac{\pi}{2}\right)$. Then the value of $\theta$ such that the sum of intercepts on axes made by this tangent is minimum is equal to ..... .

Similar Questions

The equation of the ellipse whose vertices are $(\pm 5, 0)$ and foci are $(\pm 4, 0)$ is

$S$ and $T$ are the foci of an ellipse and $B$ is the end point of the minor axis. If $\triangle STB$ is an equilateral triangle,the eccentricity of the ellipse is

The extremities of the latera recta of the ellipses $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ $(a > b)$ having a given major axis $2a$ lie on:

The locus of the feet of the perpendiculars from either focus of the ellipse $(x - y + 1)^2 + (2x + 2y - 6)^2 = 20$ onto any tangent is:

$A$ point moves such that the sum of its distances from $(ae, 0)$ and $(-ae, 0)$ is $2a$. Then the equation to its locus,where $b^2 = a^2(1 - e^2)$,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module