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

`(sqrt(1-y^(2))-log abs((1+sqrt(1-y^(2)))/(1-sqrt(1-y^(2))))= pm x+c)`

Explanation

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`