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Question

The x-intercept of the tangent to a curve is twice the ordinate of the point of contact. The equation of the curve through the point (1,1) is

A
y=eyx2y
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B
x=e2xy2x
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C
y=e2yx4x
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D
x=exy2x
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Solution

The correct option is A y=eyx2y
Equation of tangent is Yy=dydx(Xx)
for X-intercept Y=0X=xydxdy
Now, xydxdy=2y
dydx=yx2y
Put y=vx
v+xdvdx=v12v
xdvdx=2v212v
12v2v2dv=dxx
12vlogv=logx+c
x2ylogyx=logx+c
at (1,1) c=12
x2ylogyx=logx12
y=e2y2x4y
y=eyx2y

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