Question

If \(x\) and \(y\) are connected parametrically by the equation,without eliminating the parameter,find \(\frac{dy}{dx}\).
\(x=a\,secθ,y=b\,tanθ\)

Answer 1

The correct answer is \(\frac{b}{a}cosecθ\)
The given equations are \(x=asecθ,y=btanθ\)
Then,\(\frac{dx}{dθ}=a.\frac{d}{dθ}(secθ)=a\,secθ\,tanθ\)
\(\frac{dy}{dθ}=b.\frac{d}{dθ}(tan\,θ)=b\,sec2θ\)
\(∴\frac{dy}{dx}=\frac{(\frac{dy}{dθ})}{(\frac{dx}{dθ})}=\frac{b\,sec^2θ}{a\,secθ\,tanθ}\)
\(=\frac{b}{a}secθ\,cotθ=\frac{b\,cosθ}{a\,cosθ\,sinθ}=\frac{b}{a}\times \frac{1}{sinθ}=\frac{b}{a}cosecθ\)