The locus of the point of intersection of the lines √2x-y+4√2k=0 and √2kx+ky-4√2=0 (k is any non-zero real parameter) is

Correct answer: (C) a hyperbola with length of its transverse axis 8√2. Given equation of lines are √2x-y+4√2k=0 ...1 and √2kx+ky-4√2=0 ...2 Eliminating k from 1 and 2, ⇒√2x+y(√2x-y)/(-4√2)=4√2 ⇒2x^2-y^2=-32 ⇒y^232-x^216=1, which is a hyperbola. ⇒e=√(1+1632)=√32 and length of transverse axis =8√2.

Q48 / 90 · JEE Main 2018 (16 Apr Online) · Mathematics

The locus of the point of intersection of the lines $\sqrt{2} x - y + 4 \sqrt{2} k = 0$ and $\sqrt{2} k x + k y - 4 \sqrt{2} = 0$ ( $k$ is any non-zero real parameter) is

A
an ellipse whose eccentricity is $\frac{1}{\sqrt{3}}$
B
a hyperbola whose eccentricity is $\sqrt{3}$
Correct answer
C
a hyperbola with length of its transverse axis $8 \sqrt{2}$
D
an ellipse with length of its major axis $8 \sqrt{2}$

Solution

Given equation of lines are

\sqrt{2} x - y + 4 \sqrt{2} k = 0 . . . (1)

and $\sqrt{2} k x + k y - 4 \sqrt{2} = 0 . . . (2)$

Eliminating $k$ from $(1)$ and $(2)$ ,

$\Rightarrow (\sqrt{2} x + y) (\frac{\sqrt{2} x - y}{- 4 \sqrt{2}}) = 4 \sqrt{2}$

$\Rightarrow 2 x^{2} - y^{2} = - 32$

$\Rightarrow \frac{y^{2}}{32} - \frac{x^{2}}{16} = 1$ , which is a hyperbola.

$\Rightarrow e = \sqrt{1 + \frac{16}{32}} = \sqrt{\frac{3}{2}}$ and length of transverse axis

$= 8 \sqrt{2}$ .

Hence, the correct answer is option: 'C'.

The locus of the point of intersection of the lines 2x-y+42k=0 and 2kx+ky-42=0 (k is any non-zero real parameter) is

Solution

Relevant Questions

The locus of the point of intersection of the lines $\sqrt{2} x - y + 4 \sqrt{2} k = 0$ and $\sqrt{2} k x + k y - 4 \sqrt{2} = 0$ ( $k$ is any non-zero real parameter) is