Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • The vertices of a hyperbola are <img alt="( \pm 6,0)" src="https://entrancecorner.oncodecogs.com/gif.latex?%28%20%5Cpm%20

The vertices of a hyperbola Hare ( \pm 6,0) and its eccentricity is \frac{\sqrt{5}}{2}.Let \mathrm{N}  be the normal to \mathrm{H} at a point in the first quadrant and parallel to the line \sqrt{2} x+y=2 \sqrt{2}.  
If d  is the length of the line segment of Nbetween H and the y-axis then d^{2} is equal to  

Option: 1

216


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer




\mathrm{H}: \frac{\mathrm{x}^{2}}{36}-\frac{\mathrm{y}^{2}}{9}=1
Equation of normal is \quad 6 x \cos \theta+3 y \cot \theta=45
                             M =-2 \sin \theta=-\sqrt{2}
                              \theta =\pi / 4

Equation of normal is \quad \sqrt{2} x+y=15
                                  \mathrm{P}(\operatorname{asec} \theta, b \tan \theta)
                                  \mathrm{P}(6 \sqrt{2}, 3), \mathrm{k}(0,15)
                                  \mathrm{d}^{2}=216 
                               

Posted by

jitender.kumar

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

JEE Main 2025: 11 from Kota coaching centres among 24 students with 100 percentile
anam-khan

Low JEE Main rank? Explore alternative paths for engineering admission
anam-khan

Gender-gap in JEE Main 2025: Female participation remains around 31%, overall pool drops by 7.09%

Latest Question