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Question

The locus of the point of intersection of the line, √2x−y+4√2k=0 and √2kx+ky+4√20 (k is any non-zero real parameter), is

A
an ellipse whose eccentricity is 13.
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B
an ellipse with length of its major axis 82
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C
a hyperbola whose eccentricity is 3
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D
a hyperbola with length of its transverse axis 82
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Solution

The correct option is A a hyperbola whose eccentricity is √3√2x−y+4√2k=0⇒√2x−y=−4√2k..........(1)√2kx+ky+4√2=0⇒√2kx+ky=−4√2..........(2)Multiplying eqn(1)&(2), we have(√2x−y)(√2kx+ky)=(−4√2k)×(−4√2)⇒2kx2+√2kxy−√2kxy−ky2=32k⇒2kx232k−ky232k=1⇒x216−y232=1..........(3)Eqn(3) is the standard equation of hyperbola, wherea2=16b2=32As we know that, eccentricity of a hyperbola is given by-e=√1+b2a2⇒e=√1+3216⇒e=√1+2=√3Hence, the locus of the intersection of the given line is a hyperbola whose eccentricity is √3.

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