We have, x = \(a cos^3\theta\), y = \(a sin^3\theta\)
\(\implies\) \(dx\over d\theta\) = \(-3 a cos^2\theta sin\theta\), \(dy\over d\theta\) = \(3 a sin^2\theta cos\theta\)
Now, \(dy\over dx\) = \(dy/d\theta\over dx/d\theta\)
\(\implies\) \(dy\over dx\) = -\(tan\theta\)
\(\therefore\) Slope of the normal at any point on the curve = \(-1\over dy/dx\) = \(cot\theta\)
Hence, the slope of the normal at \(\theta\) = \(\pi\over 4\) = \(cot\pi\over 4\) = 1.
Find the slope of normal to the curve x = 1 – \(a sin\theta\), y = \(b cos^2\theta\) at \(\theta\) = \(\pi\over 2\).
Show that the tangents to the curve y = \(2x^3 – 3\) at the points where x =2 and x = -2 are parallel.
Find the equations of the tangent and the normal at the point ‘t’ on the curve x = \(a sin^3 t\), y = \(b cos^3 t\).
Find the equation of the normal to the curve y = \(2x^2 + 3 sin x\) at x = 0.