Q: Normal to x2/3 + y2/3 = a2/3 in the form y cosθ – x sinθ = a cos2θ ; Where θ is the angle which the normal makes with the x – axis .
Solution : x2/3 + y2/3 = a2/3
Differentiating with respect to x
$\displaystyle \frac{2}{3} x^{-1/3} + \frac{2}{3} y^{-1/3} \frac{dy}{dx} = 0 $
$\displaystyle \frac{dy}{dx} = – \frac{x^{-1/3}}{y^{-1/3}} = – \frac{y^{1/3}}{x^{1/3}}$
Hence Slope of normal is
$\displaystyle = \frac{x^{1/3}}{y^{1/3}} = tan\theta $ ; Since it is given that normal makes an angle θ with x -axis .
$\displaystyle = \frac{x^{1/3}}{y^{1/3}} = \frac{sin\theta}{cos\theta} $
$\displaystyle = \frac{x^{1/3}}{sin\theta} = \frac{y^{1/3}}{cos\theta} = [\frac{x^{2/3} + y^{2/3}}{sin^2 \theta + cos^2 \theta}]^{1/2} = a^{1/3}$
Hence , x = a sin3θ , y = a cos3θ
Therefore Equation of normal whose slope is tanθ will be
$\displaystyle y – a cos^3 \theta = \frac{sin\theta}{cos\theta} (x – a sin^3 \theta) $
$\displaystyle y cos\theta – a cos^4 \theta = x sin\theta – a sin^4 \theta) $
$\displaystyle y cos\theta – x sin\theta = a ( cos^4 \theta = – sin^4 \theta ) $
$\displaystyle y cos\theta – x sin\theta = a cos 2 \theta $