The equation of the normal to the curve x = a cos3 $\theta$, y = a sin3 $\theta$ at the point $\theta$ = $\pi$/4 is __________ .

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x = y \text { Here,} x = a \cos^3 $\theta$ \text { and } y = a \sin^3 $\theta$ dx/d$\theta$ = - 3a \cos^2 $\theta$ \sin $\theta$ \text { and } dy/d$\theta$ = 3a \sin^2 $\theta$ \cos $\theta$ \therefore dy/dx = dy/d$\theta$/dx/d$\theta$ = \frac{3a \sin^2 $\theta$ \cos $\theta$}{- 3a \cos^2 $\theta$ \sin $\theta$} = - \tan $\theta$ \text { Now,} \text { Slope of the tangent } = \left( dy/dx \right)_$\theta$ = $\pi$/4 =-\text { tan}$\pi$/4=-1 \left( x_1 , y_1 \right) = \left( a \cos^3 $\pi$/4, a \sin^3 $\pi$/4 \right) = \left( a/2$\sqrt$(2), a/2$\sqrt$(2) \right) \therefore \text { Equation of the normal } = y - y_1 = - 1/m \left( x - x_1 \right) \Rightarrow y - a/2$\sqrt$(2) = 1 \left( x - a/2$\sqrt$(2) \right) \Rightarrow x = y

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