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Question

The locus of the point of intersection of the line, 2xy+42k=0 and 2kx+ky+420 (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
2xy+42k=0
2xy=42k..........(1)
2kx+ky+42=0
2kx+ky=42..........(2)
Multiplying eqn(1)&(2), we have
(2xy)(2kx+ky)=(42k)×(42)
2kx2+2kxy2kxyky2=32k
2kx232kky232k=1
x216y232=1..........(3)
Eqn(3) is the standard equation of hyperbola, where
a2=16
b2=32
As we know that, eccentricity of a hyperbola is given by-
e=1+b2a2
e=1+3216
e=1+2=3
Hence, the locus of the intersection of the given line is a hyperbola whose eccentricity is 3.

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