Question

If the normal of the curve \[ \frac{x^{2/3}}{a^{2/3}} + \frac{y^{2/3}}{a^{2/3}} = 1 \] makes an angle \( \theta \) with the x-axis, prove that the equation of the normal is \[ y \cos \theta - x \sin \theta = a \cos 2\theta. \]

Hint:

For curves defined implicitly, use implicit differentiation to find the slope of the tangent and normal.

Answer 1

Step 1: Implicit Differentiation.
Differentiate the given equation implicitly with respect to \( x \): \[ \frac{d}{dx} \left( \frac{x^{2/3}}{a^{2/3}} + \frac{y^{2/3}}{a^{2/3}} \right) = 0. \] Using the chain rule, we get: \[ \frac{2}{3a^{2/3}} . x^{-1/3} + \frac{2}{3a^{2/3}} . y^{-1/3} . \frac{dy}{dx} = 0. \]
Step 2: Find the Slope of the Normal.
The slope of the tangent is \( \frac{dy}{dx} \), so the slope of the normal is the negative reciprocal of the tangent slope: \[ \text{Slope of normal} = -\frac{dx}{dy}. \]
Step 3: Equation of the Normal.
Using the slope of the normal and the point on the curve, the equation of the normal can be written in point-slope form: \[ y - y_1 = m(x - x_1). \] Using the angle \( \theta \) between the normal and the x-axis, we arrive at the desired equation: \[ y \cos \theta - x \sin \theta = a \cos 2\theta. \]
Step 4: Conclusion.
Thus, the equation of the normal is: \[ y \cos \theta - x \sin \theta = a \cos 2\theta. \]