Question

If x and y are connected parametrically by the equation, without eliminating the parameter, find dy/dx x=asecθ , y=btanθ

Answer 1

Given, x = asecθ and y = btanθ

x = asecθ
now differentiate x with respect to θ,
dx/dθ = a. d(secθ)/dθ = a (secθ.tanθ)
dx/dθ = asecθ.tanθ ----------(1)

y = btanθ
now differentiate y with respect to θ,
dy/dθ = b. d(tanθ)/dθ = b (sec²θ)
dy/dθ = bsec²θ ----------(2)

dividing equations (2) by (1),
{dy/dθ}/{dx/dθ}=(bsec²θ)/(asecθ.tanθ)
dy/dx = (bsecθ)/(atanθ)
= (b/a)cosecθ

hence, dy/dx = (b/a) cosecθ