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Question

Locus of the feet of the perpendicular drawn from focus of the hyperbola x2a2 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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