Find the curve for which the perpendicular from the foot of the ordinate to the tangent is of constant length.

Answer

`y^(2)+-x^(2)+c`

Explanation

Equation of the tangent at P(x,y) is
`Y-y = (dy)/(dx) (X-x)`
or `(dy)/(dx) X-Y + (y-x(dy)/(dx))=0`
Length of perpendicular QR upon the tangent from ordinate Q(x,0) is `|(x(dy)/(dx)-0+y-x(dy)/(dx))/(sqrt(1+((dy)/(dx))^(2)))|=k`
or `1+((dy)/(dx))^(2)=y^(2)/k^(2) or (dy)/(dx) = +- sqrt(y^(2)-k^(2))/k^(2)`
or `int(dy)/sqrt(y^(2)-k^(2))=+- int1/k dx`
or `log[y+sqrt(y^(2)-k^(2))]+-x/k+logc`
or `y+sqrt(y^(2)-k^(2))=ce^(+-x//k)`