The eccentric angle of the point (2, sqrt{3}) | vedclass

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

Question

(C) The given equation of the ellipse is $\frac{x^{2}}{16}+\frac{y^{2}}{4}=1$. Comparing this with $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$,we get

Vedclass · Accepted Answer

$\frac{\pi}{3}$

Vedclass · Answer

(C) The given equation of the ellipse is $\frac{x^{2}}{16}+\frac{y^{2}}{4}=1$. Comparing this with $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$,we get $a^{2}=16$ and $b^{2}=4$,so $a=4$ and $b=2$. Let $	heta$ be the eccentric angle of the point $P(2, \sqrt{3})$. The parametric coordinates of any point on the ellipse are given by $x = a \cos 	heta$ and $y = b \sin 	heta$. Substituting the values,we have $x = 4 \cos 	heta$ and $y = 2 \sin 	heta$. Given the point $(2, \sqrt{3})$,we equate: $2 = 4 \cos 	heta \Rightarrow \cos 	heta = \frac{2}{4} = \frac{1}{2}$ $\sqrt{3} = 2 \sin 	heta \Rightarrow \sin 	heta = \frac{\sqrt{3}}{2}$ Since both $\sin 	heta = \frac{\sqrt{3}}{2}$ and $\cos 	heta = \frac{1}{2}$ are satisfied at $	heta = \frac{\pi}{3}$,the eccentric angle is $\frac{\pi}{3}$.

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

Similar Questions

The line $12 x \cos \theta + 5 y \sin \theta = 60$ is tangent to which of the following curves?

The area (in sq. units) of the triangle formed by the tangent and normal to the ellipse $9x^2 + 4y^2 = 72$ at the point $(2, 3)$ with the $X$-axis is

If $CP$ and $CD$ are a pair of semi-conjugate diameters of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$,then $CP^{2}+CD^{2}=$

The eccentricity of an ellipse,with its centre as origin,is $1/2$. If one of the directrices is $x=4$,then the equation of the ellipse is given by

Vedclass Test Series

Exam Paper Generator

Online Exam Module