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Question

Let τ denote a curve y=y(x) which is in the first quadrant and let the point (1,0) lie on it. Let the tangent to τ at a point P intersect the y-axis at Yp. If PYp has length 1 for each point P on τ, then which of the following options is /are correct?


A
y=loge1+1x2x1x2
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B
xy+1x2=0
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C
y=loge1+1x2x+1x2
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D
xy1x2=0
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Solution

The correct option is B xy+1x2=0
Equation of tangent,
(yy1)=dy1dx1(xx1)
Coordinate of Yp(0,y1x1dy1dx1)
Distance of PYp=(x1)2+(y1y1+x1dy1dx1)2)=1
Generalising slope of line
dydx=±1x2xxy1x2=0
dy=±1x2xdx
dydx cannot be positive i.e f(x) cannot be increasing in first quadrant x(0,1)
Intregrating both side
y=1x2xdx
Let 1x2=t2xdx=tdt
y=t2t21dt
y=1dt+dt1t2dx
y=t+12loge|1+t1t|+c
y=1x2+12loge|1+1x211x2|+c
Point (1,0) lies on curve τ, c=0
y=1x2+12loge1+1x211x2
y=loge1+1x2x1x2



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