The perpendicular from the origin to the tangent at any point on a curve is equal to the abscissa of the point of contact. Also curve passes through the point (1,1). Then the length of intercept of the curve on the x-axis is__________

Answer

2

Explanation

Equation of tangent is `X(dy)/(dx)-Y-X(dy)/(dx)+y=0`

`Y-y = (dy)/(dx) (X-x)`
According to question,
`|(0-0-x(dy)/(dx)+y)/(sqrt((dy)/(dx))^(2)+1)|=x`
`therefore |(x(dy)/(dx)-y)/(sqrt((dx)/(dy))^(2)+1)|=x`
or `(x(dy)/(dx)-y)^(2)=x^(2)(1+((dy)/(dx))^(2))`
or `x^(2)((dy)/(dx))^(2)+y^(2)-2xy(dy)/(dx)=x^(2)+x^(2)((dy)/(dx))^(2)`
or `(y^(2)-x^(2))/(2xy)=(dy)/(dx)` (Homogeneous(1))
Put `y=vx` in (1). Then
`v+x(dv)/(dx) = (v^(2)-1)/(2v)`
`int(2v)/(v^(2)+1)dv=-int(dx)/x`
`"ln "(v^(2)+1)=-"ln "x+"ln "c`
`v^(2)+1=c/x`
It passes through (1,1). Then c=2.
For intercept of curve on x-axis, put y=0
We have `x^(2)-2=0` or `x=0,2`
Hence, length of intercept is 2.