Question

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. 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: (D)

Given:
$\sqrt{2} x-y+4 \sqrt{2} k=0$
$\sqrt{2} k x+k y-4 \sqrt{2}=0$
Now, eliminating $k$ from Eq. (2) by putting value of $k$ from Eq. (1), we get
$(\sqrt{2} x+y)\left(\frac{\sqrt{2 x}-y}{-4 \sqrt{2}}\right)=4 \sqrt{2} $
$2 x^{2}-y^{2}=-32 $
$\frac{y^{2}}{32}-\frac{x^{2}}{16}=1$
The above equation represents the hyperbola.
So, eccentricity e of this hyperbola is $e=\sqrt{1+\frac{16}{32}}=\sqrt{\frac{3}{2}}$
Length of transverse axis $=8 \sqrt{2}$