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,$\frac{dx}{dθ}=a.\frac{d}{dθ}(secθ)=a\,secθ\,tanθ$
$\frac{dy}{dθ}=b.\frac{d}{dθ}(tan\,θ)=b\,sec2θ$
$∴\frac{dy}{dx}=\frac{(\frac{dy}{dθ})}{(\frac{dx}{dθ})}=\frac{b\,sec^2θ}{a\,secθ\,tanθ}$
$=\frac{b}{a}secθ\,cotθ=\frac{b\,cosθ}{a\,cosθ\,sinθ}=\frac{b}{a}\times \frac{1}{sinθ}=\frac{b}{a}cosecθ$