Browse Questions

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

Toolbox:
• 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.