Question

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

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

Answer 1

Answer: (C)

a hyperbola whose eccentricity is $\sqrt{3}$

$\sqrt{2}x-y+4\sqrt{2}k=0$

$\Rightarrow \sqrt{2}x-y=-4\sqrt{2}k..........\left(1\right)$

$\sqrt{2}kx+ky+4\sqrt{2}=0$

$\Rightarrow \sqrt{2}kx+ky=-4\sqrt{2}..........\left(2\right)$

Multiplying ${eq}^n\left(1\right)\&\left(2\right)$, we have

$(\sqrt{2}x-y)(\sqrt{2}kx+ky)=(-4\sqrt{2}k)\times (-4\sqrt{2})$

$\Rightarrow 2k{x}^2+\sqrt{2}kxy-\sqrt{2}kxy-k{y}^2=32k$

$\Rightarrow \frac{2k{x}^2}{32k}-\frac{k{y}^2}{32k}=1$

$\Rightarrow \frac{x^2}{16}-\frac{y^2}{32}=1..........\left(3\right)$

${Eq}^n\left(3\right)$ is the standard equation of hyperbola, where

$a^2=16$

$b^2=32$

As we know that, eccentricity of a hyperbola is given by-

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

$\Rightarrow e=\sqrt{1+\frac{32}{16}}$

$\Rightarrow e=\sqrt{1+2}=\sqrt{3}$

Hence, the locus of the intersection of the given line is a hyperbola whose eccentricity is $\sqrt{3}$.