JEE Main On 8 April 2017 Question 17
Question: The locus of the point of intersection of the straight lines, $ tx-2y-3t=0 $ $ x-2ty+3=0 $ $ (t\in R), $ is: [JEE Online 08-04-2017]
Options:
A) A hyperbola with the length of conjugate axis 3
B) a hyperbola with eccentricity $ \sqrt{5} $
C) an ellipse with the length of major axis 6
D) an ellipse with eccentricity $ \frac{2}{\sqrt{5}} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ tx-2y-3t=0 $
$ x-2ty+3=0 $
$ \frac{ \begin{aligned} & tx-2y-3t=0 \\ & tx-2t^{2}y\underline{+}3t=0 \\ \end{aligned}}{y(2t^{2}-2)=6t} $
$ \frac{ \begin{aligned} & t^{2}x-2ty-3t^{2}=0 \\ & -x-2ty\underline{+}3=0 \\ \end{aligned}}{(t^{2}-1)x=(3t^{2}+1)} $
$ y\frac{6t}{2t^{2}-2}=\frac{3t}{t^{2}-1} $
$ x=-3\sec 2\theta $
$ 2y=3(-tan2\theta ) $
$ {{\sec }^{2}}2\theta -{{\tan }^{2}}2\theta =1 $
$ \frac{x^{2}}{9}-\frac{y^{2}}{9/4}=1 $
$ a^{2}=9; $
$ b^{2}=9/4 $
$ \lambda (T.A)=6; $
$ e^{2}=1+\frac{9/4}{9}=1+\frac{1}{4};e,e=\frac{\sqrt{5}}{2} $
