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

Solution:

The slope of a line is nothing but the change in y coordinate with respect to the change in x coordinate of that line.

The equation of the given curve is

ay² = x³.

On differentiating with respect to x , we have:

2ay dy/dx = 3x²

⇒ dy / dx

= (3x²)/2ay

The slope of a tangent to the curve at (am², am³) is

dy/dx]_{(am², am³)}

= 3 (am²)²/2a (am³)

= 3 (a²m⁴)/2 a²m³

= 3m/2

Therefore, the slope of normal at (am², am³) is

- 1/slope of the tangent at (am², am³)

= - 1/(3m / 2)

= - 2/3m

Hence, the equation of the normal at (am², am³) is given by,

y - am³ = - 2/3m (x - am²)

⇒ 3my - 3am⁴

= - 2x + 2am²

⇒ 2x +3my - am² (2 + 3m²) = 0

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NCERT Solutions Class 12 Maths - Chapter 6 Exercise 6.3 Question 20

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

Summary:

The equation of the normal at the point (am², am³) for the curve ay² = x³ is  2x +3my - am² (2 + 3m²) = 0