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Show that the equation of the normal to the curve $x=a\cos^{3}\theta ; y=a\sin^{3}\theta$ at $\;'\theta'$ is$\; x\cos\theta-y\cos\theta=a\cos 2\theta.$

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  • If $y=f(x)$ then $\large\frac{dy}{dx}$$=f'(x)$ is the rate of change of $y$ w.r.t $x$
  • $\large\frac{dy}{dx_{(x_1,y_1)}}$ is the slope of the tangent to the curve at the point $(x_1,y_1) $ on the curve. It is the slope of the curve at that point.
  • The normal at a point $(x_1,y_1)$ on $y=f(x)$ is perpendicular to the tangent at $(x_1,y_1)$
  • If $x=f(t)$ and $y= y(t)$ then $\large\frac{dy}{dx}=\large\frac{\Large\frac{dx}{dt}}{\Large\frac{dy}{dt}}$
$x= a \cos ^3 \theta,y=a \sin ^3 \theta$
Step 1:
Differentiating the above with respect to $\theta$
$\large\frac{dx}{d \theta}$$=3 a \cos ^2 \theta (-\sin \theta)=-3a \cos ^2 \theta \sin \theta$
$\large\frac{dy}{d \theta}$$=3 a \sin ^2 \theta \cos \theta $
$\large\frac{dy}{dx}=\frac{- \sin \theta}{ \cos \theta} $
Step 2:
The slope of the normal at '$\theta$' s $\large\frac{-1}{\Large\frac{dx}{dy}}=\large\frac{\cos \theta}{\sin \theta}$
Equation of the normal at '$\theta$' $\infty$
$y- a \sin ^3 \theta=\large\frac{\cos \theta}{\sin \theta}$$(x -a \cos ^3 \theta)$
$y \sin \theta-a \sin ^4 \theta= x \cos \theta -a \cos ^4 \theta$
Therefore $x \cos \theta-y \sin \theta= a( \cos^4 \theta - \sin ^4 \theta)$
$x \cos \theta-y \sin \theta=a( \cos ^2 \theta-\sin ^2 \theta)(\cos ^2 \theta+\sin ^2 \theta)$
$ x \cos \theta - y \sin \theta = a \cos 2 \theta $ is the required equation.
answered Jul 24, 2013 by meena.p

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