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Question

The equation of a curve passing through (1,0) for which the product of the abscissa of a point P and the intercept made by a normal at P on the x-axis equals twice the square of the distance of the point P from the origin O, is

A
x2+y2=x4
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B
x2+y2=2x4
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C
x2+y2=4x4
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D
x2+y2=8x4
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Solution

The correct option is A x2+y2=x4
Tangent at point P is Yy=1m(Xx)where m=dydx.
Let Y=0. Then X=my+x

According to question, x(my+x)=2(x2+y2)
or dydx=x2+2y2xy (homogeneous)
Putting y=vx, we get
v+xdvdx=1+2v2v
or xdvdx=1+2v2vv=1+v2v
or v dv1+v2=dxx
or 12log(1+v2)=log x+log c, c>0
or x2+y2=cx4
Also, it passes through (1, 0). Then c=1.

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