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        The locus of foot of perpendicular from the center of the hyperbola xy=c^2 on a variable tangent is                                  (1){x^2-y^2}=4c^2xy                         (2){x^2+y^2}=2c^2xy                           (3){x^2+y^2}=4c^2xy                           (4){x^2+y^2}=4c^2xy{plese explen}
8 years ago

Answers : (1)

Nishant Vora
IIT Patna
askIITians Faculty
2467 Points
							Hi student,
we have rectangular hyperbola
xy=c2

take a general point on this in parametric P(ct, c/t)
Now tangent at point P is T=0
xy1 + yx1 = 2c2
So in parametric
x(c/t) + y(ct) = 2c2
x/t + ty – 2c = 0….….…...(1)

Now centre is O(0,0)
LET foot of Perpendiculer be F(h,k)

So OF \perp tangent
So product of slopes = -1
k/h * (-1/t2)= -1
hence k/h= t2 ….…..(2)

Also prependicular distance OF is \left | \frac{2c }{\sqrt{t^2 + (\frac{1}{t})^2}} \right | = \sqrt{h^2 + k^2}

Now Put t2=k/h in above you will get the answer

I hope the solution is clear
4 years ago
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