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Question

# Locus of the feet of the perpendicular drawn from focus of the hyperbola x2a2 âˆ’ y2b2 =1 upon any tangent is _____

A

x2 + y2 = 2a2

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B

x2 + y2 = a2

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C

x2 - y2 = a2

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D

x2 - y2 = 2a2

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Solution

## The correct option is B x2 + y2 = a2 Equation of the given hyperbola is x2a2 − y2b2 =1 Let slope of the tangent be m Equation of the tangent to the hyperbola is x2a2 − y2b2 =1 y = mx ± √a2m2 − b2 Let coordinates of foot of perpendicular on tangent be (h,k) substituting (h,k) in the equation of tangentg and squaring both sides (k − mh)2 = a2m2 − b2 - - - - - - (1) Slope of the foot of perpendicular is −1slopeoftangent = −1m Equation of feet of perpendicular drawn from the focus is s(± ae,0) y − 0 = −1m(x± ae) - - -- - - - - (2) y = −1m(x± ae) Equation (2) also passes through (h,k) k = −1m(h± ae) km + h = ± ae Squaring on both sides (km + h)2 = a2e2 = a2 + b2 = a2 + b2 - - - -- - - (3) { e2 = 1 + b2a2} Adding equation (1) & (3) so that we can eliminate (m) (k − mh)2 + (km + h)2 = a2m2 − b2 + a2 + b2 h2 + k2 + m2 (h2 + k2) = a2(1+m2) (h2 + k2)(1 + m2) = a2(1 + m2) h2 + k2 − a2 For required locus x2 + y2 = a2 so the locus the feet of the perpendicular drawn from locus of the hyperbola x2a2 − y2b2= 1 upon any tangent is x2 + y2 = a2 i.e.,auxiliary circle. option B is correct.

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