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
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
Difficult
View SolutionFor which of the following curves is the line $x+\sqrt{3} y=2 \sqrt{3}$ a tangent at the point $\left(\frac{3 \sqrt{3}}{2}, \frac{1}{2}\right)$?
Tangents are drawn to the ellipse $\frac{x^2}{9}+\frac{y^2}{5}=1$ at all the ends of its latus rectum. The area of the quadrilateral so formed (in sq. units) is
DifficultTS EAMCET 2023
View SolutionThe eccentricity of the curve represented by $x = 3(\cos t + \sin t)$ and $y = 4(\cos t - \sin t)$ is
$A$ line passing through the point $P(\sqrt{5}, \sqrt{5})$ intersects the ellipse $\frac{x^2}{36} + \frac{y^2}{25} = 1$ at $A$ and $B$ such that $(PA) \cdot (PB)$ is maximum. Then $5(PA^2 + PB^2)$ is equal to:
DifficultJEE MAIN 2025
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