We have, x = 1 – \(a sin\theta\), y = \(b cos^2\theta\)
\(\implies\) \(dx\over d\theta\) = \(-a cos\theta\) and \(dy\over d\theta\) = \(-2b cos\theta sin\theta\)
\(\therefore\) \(dy\over dx\) = \(dy/d\theta\over dx/d\theta\) = \(2b\over a\) \(sin\theta\)
\(\implies\) \(dy\over dx\) at \(\pi\over 2\) = \(2b\over a\)
Hence, Slope of normal at \(\theta\) = \(\pi\over 2\) = \(-a\over 2b\)
Find the slope of the normal to the curve x = \(a cos^3\theta\), y = \(a sin^3\theta\) at \(\theta\) = \(\pi\over 4\).
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.