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Question

Locus of the pole of a chord of the hyperbola which subtends a right angles at the vertex is:

A
x=a2+b2a2b2
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B
x=a2b2a2+b2
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C
x=a2+b2a+b
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D
None of these
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Solution

The correct option is D None of these

Let the equation of hyperbola be x2a2y2b2=1 ......(i)

Vertex of the parabola is (a,0)

Shifting the origin to (a,0), then equation of hyperbola becomes

(x+a)2a2y2b2=1x2a2y2b2=2xab2x2a2y2=2ab2x ......(ii)

Let the pole of hyperbola be (h,k)

Equation of polar w.r.t to (i) b2hxa2ky=a2b2

Shifting the origin to (a,0)

b2h(x+a)a2ky=a2b2b2hxa2ky=a2b2ab2hb2hxa2kya2b2ab2h=1 .........(iii)

Making the equation of hyperbola in (ii) homogeneous using equation of polar in (iii), we get

b2x2a2y2=2ab2x{b2hxa2kya2b2ab2h}b2x2a2y2+2x{b2hxa2kyah}=0

Now the chord subtend right angle at the origin i.e. at (a,0).

Therefore, coefficient of x2 + coefficient of y2=0

b2a2+2b2hah=0h(b2a2)+a(b2a2)+2b2h=0h=a(a2b2)a2+b2

Generalizing the equation, we get

x=a(a2b2)a2+b2

So, option D is correct.


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