The equations $x = a \cos \theta$ and $y = b \sin \theta$ with $a > b$ represent a conic section. Its eccentricity $e$ is given by:
- A$e^2 = \frac{a^2 + b^2}{a^2}$
- B$e^2 = \frac{a^2 + b^2}{b^2}$
- C$e^2 = \frac{a^2 - b^2}{a^2}$
- D$e^2 = \frac{a^2 - b^2}{b^2}$
Explore More
Similar Questions
Let $S$ and $S^{\prime}$ be the foci of the ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$ and $P(\alpha, \beta)$ be a point on the ellipse in the first quadrant. If $(SP)^2+(S^{\prime}P)^2-SP \cdot S^{\prime}P=37$,then $\alpha^2+\beta^2$ is equal to:
DifficultJEE MAIN 2026
View SolutionThe angle between the tangents drawn from a point $(-3, 2)$ to the ellipse $4x^2 + 9y^2 - 36 = 0$ is
Let $P$ be an arbitrary point on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ where $a > b > 0$. Suppose $F_1$ and $F_2$ are the foci of the ellipse. The locus of the centroid of the $\triangle P F_1 F_2$ as $P$ moves on the ellipse is
The minimum area of a triangle formed by any tangent to the ellipse $\frac{x^2}{16} + \frac{y^2}{81} = 1$ and the coordinate axes is
DifficultJEE MAIN 2014
View Solution$A$ focus of an ellipse is at the origin. The directrix is the line $x = 4$ and the eccentricity is $\frac{1}{2}$. Then the length of the semi-major axis is
MediumAIEEE 2008
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo