A tangent to the ellipse frac{x^2}{a^2} + fra | vedclass

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

Question

(A) The equation of the tangent to the ellipse at $P(a \cos 	heta, b \sin 	heta)$ is given by $\frac{x}{a} \cos 	heta + \frac{y}{b} \sin 	heta = 1

Vedclass · Accepted Answer

$ab$

Vedclass · Answer

(A) The equation of the tangent to the ellipse at $P(a \cos 	heta, b \sin 	heta)$ is given by $\frac{x}{a} \cos 	heta + \frac{y}{b} \sin 	heta = 1$. Setting $y = 0$,we get the $x$-intercept $A = (\frac{a}{\cos 	heta}, 0)$. Setting $x = 0$,we get the $y$-intercept $B = (0, \frac{b}{\sin 	heta})$. The area of $\Delta OAB$ is given by $Area = \frac{1}{2} |x_A \cdot y_B| = \frac{1}{2} |\frac{a}{\cos 	heta} \cdot \frac{b}{\sin 	heta}| = \frac{ab}{|2 \sin 	heta \cos 	heta|} = \frac{ab}{|\sin 2	heta|}$. Since the minimum value of $|\sin 2	heta|$ is $1$ (when $2	heta = 90^\circ$ or $	heta = 45^\circ$),the minimum area is $ab$.

$A$ tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at point $P$ meets the coordinate axes at points $A$ and $B$. Find the minimum area of $\Delta OAB$.

Similar Questions

The eccentricity of the ellipse $9x^2 + 25y^2 = 225$ is

The locus of the point of intersection of perpendicular tangents to the ellipse $\frac{x^2}{25} + \frac{y^2}{16} = 1$ is:

The coordinates of a point,in the parametric form,on the ellipse whose foci are $(-1, 0)$ and $(7, 0)$ and eccentricity $e = \frac{1}{2}$,are

Equations of the latus rectum of the ellipse $9x^2+4y^2-18x-8y-23=0$ are:

The length of the latus rectum of the ellipse $9x^2 + 4y^2 = 1$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module