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

1.	Find the equation of the normal to the curve ay2 = x3 at the point (am2, am3).
Answer» Given as ay² = x³ Differentiate with respect to x, to get slope of tangent 2ay(dy/dx) = 3x² dy/dx = 3x²/2ay m(tangent) at (am², am³) is 3m/2 The normal is perpendicular to tangent therefore, m₁m₂ = – 1 m(normal) at (am², am³) is -2/3m The equation of normal is given by y – y₁ = m(normal)(x – x₁) y - am³ = (-2/3m)(x - am²)

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

Answer»

Given as ay² = x³

Differentiate with respect to x, to get slope of tangent 2ay(dy/dx) = 3x²

dy/dx = 3x²/2ay

m(tangent) at (am², am³) is 3m/2

The normal is perpendicular to tangent therefore, m₁m₂ = – 1

m(normal) at (am², am³) is -2/3m

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

y - am³ = (-2/3m)(x - am²)