Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

The product of the perpendiculars drawn from the foci of the ellipse , \frac{x^{2}}{9}+\frac{y^{2}}{25}=1   upon the tangent to it at the point
\left ( \frac{3}{2} ,\frac{5\sqrt{3}}{2}\right ),\; is:
  • Option 1)

    3\sqrt{3}\;

  • Option 2)

    9\;

  • Option 3)

    \frac{189}{13}\;

  • Option 4)

    18

 

Answers (1)

best_answer

As learnt in

Foci of Ellipse -

The two fixed points on the ellipse.

- wherein

 

 Product of perpendicular drawn from foci is always = b2

In ellipse \frac{x^{2}}{9}+\frac{y^{2}}{25}=1;\frac{x^{2}}{9}+\frac{y^{2}}{25}=1;\:a^{2}=25,\:b^{2}=9

Hence product =9


Option 1)

3\sqrt{3}\;

This option is incorrect.

Option 2)

9\;

This option is correct.

Option 3)

\frac{189}{13}\;

This option is incorrect.

Option 4)

18

This option is incorrect.

Posted by

prateek

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

NTA Exam Calendar 2025 Live: JEE Main, NEET, CUET schedule to be announced soon; download at nta.ac.in
anam-khan

NTA Exam Calendar 2025: JEE Main, NEET, CUET exam dates soon
anam-khan

JEE Main, NEET, CUET 2025 Dates Live: NTA to announce exam calendar soon on nta.ac.in; latest updates

Latest Question