Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • A hyperbola whose transverse axis is along the major axis of the conic   <img alt="\mathrm {\frac{x^2}{3}+\frac{y^2}{4}=4,}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm

A hyperbola whose transverse axis is along the major axis of the conic   \mathrm {\frac{x^2}{3}+\frac{y^2}{4}=4,}  and has vertices at the foci of this conic. If the eccentricity of the hyperbola is   \mathrm {\frac{3}{2}}  , then which of the following points does NOT lie on it?
 

Option: 1

(\sqrt{5}, 2 \sqrt{2})


Option: 2

(0,2)


Option: 3

(5,2 \sqrt{3})


Option: 4

(\sqrt{10}, 2 \sqrt{3})


Answers (1)

best_answer

We have,  \mathrm { \frac{x^2}{12}+\frac{y^2}{16}=1 } 
 \mathrm { \begin{aligned} & \therefore \quad e=\sqrt{1-\frac{12}{16}}=\frac{1}{2} \\ & \Rightarrow \text { Foci } \equiv(0,2) \&(0,-2) \end{aligned} }


So, transverse axis of hyperbola  \mathrm {=2 b=4 \Rightarrow b=2\: \: and \: \: a^2=b^2\left(e^2-1\right) \Rightarrow a^2=4\left(\frac{9}{4}-1\right) \Rightarrow a^2=5\\ }

\therefore \quad \text {Required equation is }\mathrm {\frac{x^2}{5}-\frac{y^2}{4}=-1}

Posted by

himanshu.meshram

View full answer

JEE Main high-scoring chapters and topics

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

Similar Questions

Latest Question