Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Which one is an end point of latus rectum of Hyperbola <img alt="\frac{\left ( x-2 \right )^{2}}{4}-\left ( y+1 \right )^{2}=1" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cfrac%7B%

Which one is an end point of latus rectum of Hyperbola \frac{\left ( x-2 \right )^{2}}{4}-\left ( y+1 \right )^{2}=1

Option: 1

\left ( 2-\sqrt5,5/2 \right )


Option: 2

\left ( 2+\sqrt5,5/2 \right )


Option: 3

\left ( 2-\sqrt5,3/2 \right )


Option: 4

\left ( 2+\sqrt5,-3/2 \right )


Answers (1)

best_answer

As we learnt

 

End point of latus rectum -

 

\left ( \pm ae, \pm \frac{b^{2}}{a}\right )

- wherein

For the Hyperbola

\frac{x^{2}}{a^{2}}- \frac {y^{2}}{b^{2}}= 1

 

Here, a^2=4,b^2=1

e=\sqrt{1+\frac{b^2}{a^2}}= \frac{\sqrt{5}}{2}

Thus ae=\sqrt5\: \: and\: \: \frac{b^2}{a}=\frac{1}{2}

and centre is (2,-1)

Thus end points are \left ( 2\pm \sqrt{5},-1\pm\frac{1}{2} \right )

Thus, \left ( 2+ \sqrt{5},\frac{-3}{2} \right ) is an end point

 

Posted by

mansi

View full answer

JEE Main high-scoring chapters and topics

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

Similar Questions