Q9  If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find dy/dx  x = a \sec \theta , y = b \ tan \theta

Answers (1)

Given equations are
x = a \sec \theta , y = b \ tan \theta
Now, differentiate both w.r.t  \theta
We get,
\frac{dx}{d\theta}=\frac{d(a\sec \theta)}{d\theta}= a\sec \theta \tan \theta
Similarly,
\frac{dy}{d\theta}=\frac{d(b\tan \theta)}{d\theta}= b\sec^2 \theta
Now, \frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}= \frac{b\sec^2 \theta}{a\sec\theta\tan \theta} = \frac{b\sec\theta}{a\tan \theta}= \frac{b\frac{1}{\cos\theta}}{a\frac{\sin \theta}{\cos \theta}} = \frac{b }{a\sin \theta} = \frac{b cosec \theta}{a}
Therefore, the answer is \frac{dy}{dx} = \frac{b cosec \theta}{a}
 

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