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Find the curve for which the perpendicul...

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

Text Solution

Verified by Experts

The correct Answer is:
`y^(2)+-x^(2)+c`

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)`
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