If the polar with respect to y2=4x touches the ellipse x2α2+y2β2=1, the locus of its pole is
A
x2α2−y2(4α2/β2)=1
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B
x2α2+β2y24a2=1
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C
α2x2+β2y2=1
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D
None of these
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Solution
The correct option is Ax2α2−y2(4α2/β2)=1 Let pole be P(h,k). thus it's polar w.r.t y2=4x is given by, T=0⇒ky=2(x+h) or y=2kx+2hk.
This line touch ellipse, x2α2+y2β2=1 Thus applying condition of tangency, c2=a2m2+b2 ⇒4h2k2=4α2k2+β2⇒h2α2−k2(4α2/β2)=1 Hence required locus of P(h,k) is x2α2−y2(4α2β2)=1