JEE Main On 16 April 2018 Question 4
Question: The locus of the point of intersection of the lines, $ \sqrt{2x}-y+4\sqrt{2k}=0 $ and $ \sqrt{2}kx+ky-4\sqrt{2}=0 $ (k is any non-zero real parameter), is? [JEE Main 16-4-2018]
Options:
A) A hyperbola with length of its transeverse axis $ 8\sqrt{2} $
B) An ellipse with length of its major axis $ 8\sqrt{2} $
C) An ellipse whose eccentricity is $ \frac{1}{\sqrt{3}} $
D) A hyperbola whose eccentricity is $ \sqrt{3} $
Show Answer
Answer:
Correct Answer: A
Solution:
Gives lines are: $ \sqrt{2x}-y+4\sqrt{2k}=0 $
$ \Rightarrow $ $ \sqrt{2x}+4\sqrt{2k}=y $ ..(i) and
$ \sqrt{2kx}+ky-4\sqrt{2}=0 $ ?.. (ii)
We have from the equations of the lines: Substituting (i) in (ii),
$ \Rightarrow 2\sqrt{2}kx+4\sqrt{2}(k^{2}-1)=0 $
$ \Rightarrow $ $ x=\frac{1(1-k^{2})}{k},y=\frac{2\sqrt{2}(1+k^{2})}{k} $
$ \Rightarrow $ $ {{( \frac{y}{4\sqrt{2}} )}^{2}}-{{( \frac{x}{4} )}^{2}}=1 $
$ \Rightarrow $ $ {{( \frac{y}{4\sqrt{2}} )}^{2}}-{{( \frac{x}{4} )}^{2}}=1 $
Locus of transverse axis $ =2\sqrt{32} $ $ =2\times 4\sqrt{2} $ $ =8\sqrt{2} $
Thus, the locus is a hyperbola with length of its transverse axis equal to $ 8\sqrt{2}. $ So option A is the correct answer.
