If $F_1$ and $F_2$ are the feet of the perpendiculars from the foci $S_1$ and $S_2$ of an ellipse $\frac{x^2}{5} + \frac{y^2}{3} = 1$ on the tangent at any point $P$ on the ellipse,then $(S_1 F_1) (S_2 F_2)$ is equal to
- A$2$
- B$3$
- C$4$
- D$5$
Explore More
Similar Questions
The eccentric angle of the point $(2, \sqrt{3})$ lying on $\frac{x^{2}}{16}+\frac{y^{2}}{4}=1$ is
An ellipse having the coordinate axes as its axes and its major axis along the $Y$-axis,passes through the point $(-3, 1)$ and has eccentricity $e = \sqrt{\frac{2}{5}}$. Then its equation is:
DifficultAP EAMCET 2018
View SolutionIf the distance between the two foci of an ellipse is equal to its minor axis,then the eccentricity of the ellipse is:
Easy
View SolutionEquations of the latus rectum of the ellipse $9x^2+4y^2-18x-8y-23=0$ are:
DifficultAP EAMCET 2006
View SolutionThe eccentricity of an ellipse having centre at the origin,axes along the coordinate axes and passing through the points $(4, -1)$ and $(-2, 2)$ is
DifficultJEE MAIN 2017
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