Question

The equation of the normal to the curve x = a cos³ θ, y = a sin³ θ at the point θ = π/4 is __________ .

Options

• x = 0

• y = 0

• x = y

• x + y = a

Answer 1

`x = y`

 

\[\text { Here,} \]

\[x = a \cos^3 \theta \text { and } y = a \sin^3 \theta\]

\[\frac{dx}{d\theta} = - 3a \cos^2 \theta \sin \theta \text { and } \frac{dy}{d\theta} = 3a \sin^2 \theta \cos \theta\]

\[ \therefore \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{3a \sin^2 \theta \cos \theta}{- 3a \cos^2 \theta \sin \theta} = - \tan \theta\]

\[\text { Now,} \]

\[\text { Slope of the tangent } = \left( \frac{dy}{dx} \right)_\theta = \frac{\pi}{4} =-\text { tan}\frac{\pi}{4}=-1\]

\[\left( x_1 , y_1 \right) = \left( a \cos^3 \frac{\pi}{4}, a \sin^3 \frac{\pi}{4} \right) = \left( \frac{a}{2\sqrt{2}}, \frac{a}{2\sqrt{2}} \right)\]

\[ \therefore \text { Equation of the normal }\]

\[ = y - y_1 = \frac{- 1}{m} \left( x - x_1 \right)\]

\[ \Rightarrow y - \frac{a}{2\sqrt{2}} = 1 \left( x - \frac{a}{2\sqrt{2}} \right)\]

\[ \Rightarrow x = y\]