The eccentric angle of the point $(2,\sqrt{3})$ lying on $\frac {x^2}{16}+\frac{y^2}{4}-1$ is _________
- $\frac {\pi}{3}$
- $\frac {\pi}{6}$
- $\frac {\pi}{4}$
- $\frac {\pi}{2}$
The Correct Option is A
Solution and Explanation
$\frac{x^{2}}{16}+\frac{y^{2}}{4}=1$ and point $P(2, \sqrt{3})$
Let $\theta$ be the eccentric angle.
The parametric coordinate of an ellipse is
$\begin{cases}
x=4 \cos \theta \\
y=2 \sin \theta
\end{cases}$...(i)
Given that, eccentric angle at $P$ is,
$2=4 \cos \theta \Rightarrow \cos \theta=\frac{1}{2}$
$\sqrt{2}=2 \sin \theta \Rightarrow \sin \theta=\frac{\sqrt{3}}{2}$
Hence, $\theta=\frac{\pi}{3}$
Top Questions on Ellipse
- Let the product of the focal distances of the point \[ \left( \sqrt{3}, \frac{1}{2} \right) \] on the ellipse \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \quad (a>b), \] be \( \frac{7}{4} \). Then the absolute difference of the eccentricities of two such ellipses is:
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- The product of perpendiculars from the two foci of the ellipse \[ \frac{x^2}{9} + \frac{y^2}{25} = 1 \] on the tangent at any point on the ellipse is:
- The product of perpendiculars from the two foci of the ellipse \[ \frac{x^2}{9} + \frac{y^2}{25} = 1 \] on the tangent at any point on the ellipse is:
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Concepts Used:
Ellipse
Ellipse Shape
An ellipse is a locus of a point that moves in such a way that its distance from a fixed point (focus) to its perpendicular distance from a fixed straight line (directrix) is constant. i.e. eccentricity(e) which is less than unity
Properties
- Ellipse has two focal points, also called foci.
- The fixed distance is called a directrix.
- The eccentricity of the ellipse lies between 0 to 1. 0≤e<1
- The total sum of each distance from the locus of an ellipse to the two focal points is constant
- Ellipse has one major axis and one minor axis and a center
Read More: Conic Section
Eccentricity of the Ellipse
The ratio of distances from the center of the ellipse from either focus to the semi-major axis of the ellipse is defined as the eccentricity of the ellipse.
The eccentricity of ellipse, e = c/a
Where c is the focal length and a is length of the semi-major axis.
Since c ≤ a the eccentricity is always greater than 1 in the case of an ellipse.
Also,
c2 = a2 – b2
Therefore, eccentricity becomes:
e = √(a2 – b2)/a
e = √[(a2 – b2)/a2] e = √[1-(b2/a2)]
Area of an ellipse
The area of an ellipse = πab, where a is the semi major axis and b is the semi minor axis.
Position of point related to Ellipse
Let the point p(x1, y1) and ellipse
(x2 / a2) + (y2 / b2) = 1
If [(x12 / a2)+ (y12 / b2) − 1)]
= 0 {on the curve}
<0{inside the curve}
>0 {outside the curve}