The lengths of the major and minor axes of an ellipse are $10$ and $8$ respectively,and its major axis lies along the $y$-axis. The equation of the ellipse,with its center at the origin,is:
- A$\frac{x^2}{25} + \frac{y^2}{16} = 1$
- B$\frac{x^2}{16} + \frac{y^2}{25} = 1$
- C$\frac{x^2}{100} + \frac{y^2}{64} = 1$
- D$\frac{x^2}{64} + \frac{y^2}{100} = 1$
Explore More
Similar Questions
Find the locus of the midpoint of the portion of the tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ intercepted between the axes.
Difficult
View SolutionLet $A_1, A_2, A_3$ be regions in the $XY$-plane defined by:
$A_1 = \{(x, y) : x^2 + 2y^2 \leq 1\}$
$A_2 = \{(x, y) : |x|^3 + 2\sqrt{2}|y|^3 \leq 1\}$
$A_3 = \{(x, y) : \max(|x|, \sqrt{2}|y|) \leq 1\}$
Then,
$A_1 = \{(x, y) : x^2 + 2y^2 \leq 1\}$
$A_2 = \{(x, y) : |x|^3 + 2\sqrt{2}|y|^3 \leq 1\}$
$A_3 = \{(x, y) : \max(|x|, \sqrt{2}|y|) \leq 1\}$
Then,
What is the equation of the ellipse with foci $(\pm 2, 0)$ and eccentricity $e = \frac{1}{2}$?
Easy
View SolutionFind the equation for the ellipse that satisfies the given conditions: Ends of major axis $(0, \pm \sqrt{5})$,ends of minor axis $(\pm 1, 0)$.
Medium
View SolutionThe distance between the foci of an ellipse is $16$ and its eccentricity is $\frac{1}{2}$. The length of the major axis of the ellipse is:
Easy
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