The eccentric angles of the extremities of th | vedclass

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

Question

(C) The coordinates of any point on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with eccentric angle $	heta$ are $(a \cos 	heta, b \sin 	he

Vedclass · Accepted Answer

$	an^{-1}\left( \pm \frac{b}{ae} ight)$

Vedclass · Answer

(C) The coordinates of any point on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with eccentric angle $	heta$ are $(a \cos 	heta, b \sin 	heta)$.The coordinates of the endpoints of the latus recta are $(ae, \pm \frac{b^2}{a})$.Equating the coordinates,we have $a \cos 	heta = ae$ and $b \sin 	heta = \pm \frac{b^2}{a}$.From $a \cos 	heta = ae$,we get $\cos 	heta = e$.From $b \sin 	heta = \pm \frac{b^2}{a}$,we get $\sin 	heta = \pm \frac{b}{a}$.Therefore,$	an 	heta = \frac{\sin 	heta}{\cos 	heta} = \frac{\pm b/a}{e} = \pm \frac{b}{ae}$.Thus,$	heta = 	an^{-1}\left( \pm \frac{b}{ae} ight)$.

The eccentric angles of the extremities of the latus recta of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ are given by

Similar Questions

If $y = mx + c$ is a tangent to the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$,then the value of $c$ is ...

The eccentricity of the ellipse $9x^2 + 5y^2 - 18x - 20y - 16 = 0$ is:

The eccentricity of an ellipse passing through $(3 \sqrt{2}, \sqrt{10})$ with foci at $(-4,0)$ and $(4,0)$ is

If the normal at the point $P(\theta)$ to the ellipse $\frac{x^2}{14} + \frac{y^2}{5} = 1$ intersects it again at the point $Q(2\theta)$,then $\cos \theta$ is equal to

Let the equations of two ellipses be $E_1: \frac{x^2}{3} + \frac{y^2}{2} = 1$ and $E_2: \frac{x^2}{16} + \frac{y^2}{b^2} = 1$. If the product of their eccentricities is $\frac{1}{2}$,then the length of the minor axis of ellipse $E_2$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module