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
Let $C$ be the centre of an ellipse and $PQ$ be a chord of it with $\angle PCQ = 90^{\circ}$. If $R$ is the point of intersection of the tangents to the ellipse at $P$ and $Q$,then $R$ lies on
DifficultAP EAMCET 2017
View SolutionFor the ellipse $4x^2 + y^2 - 8x + 2y + 1 = 0$,the focus and the equation of the directrix are respectively
Find the equation of the ellipse,whose length of the major axis is $20$ and foci are $(0, \pm 5)$.
Medium
View SolutionThe product of the lengths of the perpendiculars drawn from the two foci of the ellipse $\frac{x^2}{9}+\frac{y^2}{25}=1$ to the tangent at any point on the ellipse is
DifficultAP EAMCET 2017
View SolutionLet $e$ be the eccentricity of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. If $a=5, b=4$ and the equation of the normal drawn at one end of the latus rectum that lies in the first quadrant is $lx+my=27$,then $l+m=$
Vedclass 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