lf length of tangent at any point on th curve `y=f(x)` intercepted between the point and the x-axis is of length 1. Find the equation of the curve.

lf length of tangent at any point on th curve `y=f(x)` intercepted bet

1.	lf length of tangent at any point on th curve `y=f(x)` intercepted between the point and the x-axis is of length 1. Find the equation of the curve.
Answer» Correct Answer - `(sqrt(1-y^(2))-log abs((1+sqrt(1-y^(2)))/(1-sqrt(1-y^(2))))= pm x+c)` Since, the length of tangent = `abs(ysqrt(1+(dx/dy)^(2)))=1` `rArr y^(2) (1+((dx)/dy)^2)=1` `therefore dy/dx=pm y/(sqrt(1-y^(2)))` `rArr int (sqrt(1-y^(2)))/y dy =pm int xdx ` `rArr int (sqrt(1-y^(2)))/y dy =pm x+C` Lut `y=sin theta rArr dy = cos theta d theta` `therefore int (cos theta)/(sin theta)cdot cos theta d theta = pm x+C` `therefore int (cos^(2) theta)/(sin^(2) theta)cdot sin theta d theta = pm x+C` Again put `cos theta =t rArr -sin theta d theta =dt` `therefore -int t^(2) /(1-t^(2))dt=pm x+C` `rArr int (1-1/(1-t^(2))) dt =pm x+C` `rArr t-log abs((1=t)/(1-t))=pmx+C` `rArr sqrt(1-y^(2))-log abs((1+sqrt(1-y^(2)))/(1-sqrt(1-y^(2))))=pm x+C`

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lf length of tangent at any point on th curve `y=f(x)` intercepted between the point and the x-axis is of length 1. Find the equation of the curve.

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