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Question

Mathematics Question on Conic sections

If a circle of radius RR passes through the origin OO and intersects the coordinate axes at AA and BB, then the locus of the foot of perpendicular from OO on ABAB is :

A

(x2+y2)2=4Rx2y2(x^2 + y^2)^2 = 4Rx^2y^2

B

(x2+y2)(x+y)=R2xy(x^2 + y^2)(x+y) = R^2xy

C

(x2+y2)3=4R2x2y2(x^2 + y^2)^3 = 4R^2x^2y^2

D

(x2+y2)2=4R2x2y2(x^2 + y^2)^2 = 4R^2x^2y^2

Answer

(x2+y2)3=4R2x2y2(x^2 + y^2)^3 = 4R^2x^2y^2

Explanation

Solution

Slope of AB=hkAB = \frac{-h}{k}
Equation of ABAB is hx+ky=h2+k2hx + ky = h^2 + k^2
A(h2+k2h,0),B(0,h2+k2k)A\bigg(\frac{h^2 + k^2}{h}, 0 \bigg), B\bigg(0, \frac{h^2 + k^2}{k}\bigg)
AB=2RAB = 2R
(h2+k2)3=4R2h2k2\Rightarrow \, (h^2 + k^2)^3 = 4R^2h^2k^2
(x2+y2)3=4R2x2y2\Rightarrow \, (x^2 + y^2)^3 \, = 4R^2x^2y^2