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Question

A normal to the hyperbola, 4x29y2=36 meets the co-ordinate axes x and y at A and B, respectively. If the parallelogram OABP (O being the origin) is formed, then the locus of P is :

A
4x29y2=121
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B
4x2+9y2=121
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C
9x24y2=169
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D
9x2+4y2=169
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Solution

The correct option is C 9x24y2=169
The equation of hyperbola is 4x29y2=36
where a=3 and b=2
Let the tangent pass through point P(asecθ,btanθ)
So, point P is (3secθ,2tanθ)
Equation of tangent on hyperbola is
4x((3secθ)9y(2tanθ)=36
2secθx3tanθy=6
slope=23sinθ
slope of normal=3sinθ2
Equation of normal is
y2tanθ=3sinθ2(x3secθ)
x13secθ3y13tanθ2=1
So, co-orinates of A and B are A(133secθ,0) and B(0,132tanθ)

From the midpoint of diagonals of parallelogram OABP
mid point of AP mid point of OB
M⎜ ⎜ ⎜133secθ+h2,k2⎟ ⎟ ⎟M⎜ ⎜ ⎜0,132tanθ2⎟ ⎟ ⎟
133secθ+h2=0, 132tanθ2=k2
secθ=3h13, tanθ=2k13
secθtan2θ=1
9h21694k2169=1
9h24k2=169
Hence, 9x24y2=169

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