A tangent is drawn to the ellipse frac{x^2}{2 | vedclass

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

Question

(B) Given the ellipse $\frac{x^2}{27} + y^2 = 1$,we have $a^2 = 27$ and $b^2 = 1$,so $a = 3\sqrt{3}$ and $b = 1$.The equation of the tangent at $(a \c

Vedclass · Accepted Answer

$\frac{\pi}{6}$

Vedclass · Answer

(B) Given the ellipse $\frac{x^2}{27} + y^2 = 1$,we have $a^2 = 27$ and $b^2 = 1$,so $a = 3\sqrt{3}$ and $b = 1$.The equation of the tangent at $(a \cos 	heta, b \sin 	heta)$ is $\frac{x}{a} \cos 	heta + \frac{y}{b} \sin 	heta = 1$.Substituting the values,the equation of the tangent is $\frac{x}{3\sqrt{3}} \cos 	heta + y \sin 	heta = 1$.The $x$-intercept is $3\sqrt{3} \sec 	heta$ and the $y$-intercept is $\csc 	heta$.Let $S$ be the sum of the intercepts: $S = 3\sqrt{3} \sec 	heta + \csc 	heta$.To find the minimum,differentiate $S$ with respect to $	heta$: $\frac{dS}{d	heta} = 3\sqrt{3} \sec 	heta 	an 	heta - \csc 	heta \cot 	heta$.Setting $\frac{dS}{d	heta} = 0$,we get $3\sqrt{3} \frac{\sin 	heta}{\cos^2 	heta} = \frac{\cos 	heta}{\sin^2 	heta}$.This simplifies to $	an^3 	heta = \frac{1}{3\sqrt{3}} = (\frac{1}{\sqrt{3}})^3$.Thus,$	an 	heta = \frac{1}{\sqrt{3}}$,which implies $	heta = \frac{\pi}{6}$.Since the second derivative is positive at $	heta = \frac{\pi}{6}$,the sum of intercepts is minimum at $	heta = \frac{\pi}{6}$.

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

Similar Questions

Let $E$ be an ellipse whose axes are parallel to the coordinate axes,having its center at $(3, -4)$,one focus at $(4, -4)$,and one vertex at $(5, -4)$. If $mx - y = 4$ with $m > 0$ is a tangent to the ellipse $E$,then the value of $5m^{2}$ is equal to .....

The value of $k$ such that the line $y=2x+k$ touches the ellipse $3x^2+5y^2=15$ 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:

The latus rectum of an ellipse is $10$ and the minor axis is equal to the distance between the foci. The equation of the ellipse is

The eccentric angle of the point $(2, \sqrt{3})$ lying on $\frac{x^{2}}{16}+\frac{y^{2}}{4}=1$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module