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Question

Mathematics Question on Hyperbola

The locus of the point of intersection of the lines, 2xy+42k=0\sqrt{2} x-y + 4 \sqrt{2} \, k=0 and 2kx+ky42=0\sqrt{2} k x+k \, y-4 \sqrt{2} = 0 (k is any non-zero real parameter), is :

A

an ellipse whose eccentricity is 13\frac{1}{\sqrt{3}}

B

an ellipse with length of its major axis 828\sqrt{ 2}

C

a hyperbola whose eccentricity is 3\sqrt{3}

D

a hyperbola with length of its transverse axis 828\sqrt{ 2}

Answer

a hyperbola with length of its transverse axis 828\sqrt{ 2}

Explanation

Solution

Given:
2xy+42k=0\sqrt{2} x-y+4 \sqrt{2} k=0
2kx+ky42=0\sqrt{2} k x+k y-4 \sqrt{2}=0
Now, eliminating kk from E (2) by putting value of kk from E (1), we get
(2x+y)(2xy42)=42(\sqrt{2} x+y)\left(\frac{\sqrt{2 x}-y}{-4 \sqrt{2}}\right)=4 \sqrt{2}
2x2y2=322 x^{2}-y^{2}=-32
y232x216=1\frac{y^{2}}{32}-\frac{x^{2}}{16}=1
The above equation represents the hyperbola.
So, eccentricity e of this hyperbola is e=1+1632=32e=\sqrt{1+\frac{16}{32}}=\sqrt{\frac{3}{2}}
Length of transverse axis =82=8 \sqrt{2}