Find the equation of normal to the curve ay² = x³ at the point whose x coordinate is am².

The curve given is ay² = x³

We have to find the equation of normal at point whose x-coordinate is am²

Let us first find its y-coordinate by putting x = am² in curve equation

⇒ ay² = a³m⁶

⇒ y² = a²m⁶

⇒ y = am³

Hence the point at which we have to find equation of normal is (am², am³)

Now to write equation of normal we also need its slope

Let m_n be the slope of normal at (am², am³)

Slope of tangent at (am², am³) is given by let it be denoted by m_t

ay² = x³

Differentiate with respect to x

Put (am², am³) in to get slope at that point

As tangent and normal at the same point on curve are perpendicular to each other their product of slopes is -1

⇒ m_tm_n = -1

Writing equation of normal at point (am², am³) and having slope in slope point form is

⇒ 3my – 3am⁴ = -2x + 2am²

⇒ 2x + 3my = 3am⁴ + 2am²

Hence equation of normal to curve at point whose x-coordinate is am² is 2x + 3my = 3am⁴ + 2am²