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 focus of an ellipse is at the origin. The line $x = 4$ is a directrix and the eccentricity is $1/2$. Find the length of the major axis. (in $/3$)

If the lines joining the foci of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ (where $a > b$) to an extremity of its minor axis are inclined at an angle of $60^{\circ}$ to each other, then the eccentricity of the ellipse is:

Find the equation of the tangent to the ellipse $4x^2 + 9y^2 = 36$ at the point $(3, -2)$.

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 the distance between a focus and the corresponding directrix of an ellipse is $8$ and the eccentricity is $1/2$,then the length of the minor axis is

Vedclass Test Series

Exam Paper Generator

Online Exam Module