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If \(x\) and \(y\) are connected parametrically by the equations given in $x = a \sec \theta, y = b \tan \theta $ without eliminating the parameter, Find \(\frac{\large dy}{\large dx}\).

$\begin{array}{1 1} \frac{b}{a}\begin{bmatrix}sec\theta\end{bmatrix} \\ \frac{b}{a}\begin{bmatrix}cos\theta\end{bmatrix} \\ a\begin{bmatrix}cosec\theta\end{bmatrix} \\ \frac{b}{a}\begin{bmatrix}cosec\theta\end{bmatrix} \\ \end{array} $

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  • By chain rule we have $\large\frac{dy}{dx}=\frac{dy}{d\theta}$$\times\large\frac{d\theta}{dx}$
Step 1:
$x=a\sec\theta$
Differentiating with respect to $\theta$
$\large\frac{dx}{d\theta}$$=a\sec\theta.\tan \theta$
Step 2:
$y=b\tan \theta$
Differentiating with respect to $\theta$
$\large\frac{dy}{d\theta}$$=b\sec^2\theta$
Step 3:
$\large\frac{dy}{dx}=\frac{dy}{d\theta}$$\times\large\frac{d\theta}{dx}$
$\quad\;\;=b\sec^2\theta\times \large\frac{1}{a\sec\theta\tan\theta}$
$\quad\;\;=\large\frac{b\sec\theta}{a\tan\theta}$
$\quad\;\;=\large\frac{b\big(\Large\frac{1}{\cos \theta}\big)}{a\big(\Large\frac{\sin\theta}{\cos \theta}\big)}$
$\quad\;\;=\large\frac{b}{a}\begin{bmatrix}\large\frac{1}{\cos \theta}\normalsize\times \large\frac{\cos\theta}{\sin\theta}\end{bmatrix}$
$\quad\;\;=\large\frac{b}{a}\begin{bmatrix}\large\frac{1}{\sin \theta}\end{bmatrix}$
$\quad\;\;=\large\frac{b}{a}$$\begin{bmatrix}cosec\theta\end{bmatrix}$
$\large\frac{dy}{dx}=\large\frac{b}{a}$$\begin{bmatrix}cosec\theta\end{bmatrix}$
answered May 10, 2013 by sreemathi.v
 

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