Find the equation of the normal to the curve ay2 = x3 at the point (am

1.	Find the equation of the normal to the curve ay2 = x3 at the point (am2, am3).
Answer» finding the slope of the tangent by differentiating the curve \(2ay\frac{dy}{dx}=3x^2\) \(\frac{dy}{dx}=\frac{3x^2}{2ay}\) m(tangent) at (am², am³) is \(\frac{3m}{2}\) normal is perpendicular to tangent so, m₁m₂ = – 1 m(normal) at (am², am³) is \(-\frac{2}{3m}\) equation of normal is given by y – y₁ = m(normal)(x – x₁) \(y-am^3=-\frac{2}{3m}(x-am^2)\)

Find the equation of the normal to the curve ay2 = x3 at the point (am2, am3).

Answer»

finding the slope of the tangent by differentiating the curve

\(2ay\frac{dy}{dx}=3x^2\)

\(\frac{dy}{dx}=\frac{3x^2}{2ay}\)

m(tangent) at (am², am³) is \(\frac{3m}{2}\)

normal is perpendicular to tangent so, m₁m₂ = – 1

m(normal) at (am², am³) is \(-\frac{2}{3m}\)

equation of normal is given by y – y₁ = m(normal)(x – x₁)

\(y-am^3=-\frac{2}{3m}(x-am^2)\)