# Find the equation of the normal at the point (am^{2}, am^{3}) for the curve ay^{2} = x^{3}

**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

ay^{2} = x^{3}.

On differentiating with respect to x , we have:

2ay dy/dx = 3x^{2}

⇒ dy / dx

= (3x^{2})/2ay

The slope of a tangent to the curve at (am^{2}, am^{3}) is

dy/dx]_{(am2, am3)}

= 3 (am^{2})^{2}/2a (am^{3})

= 3 (a^{2}m^{4})/2 a^{2}m^{3}

= 3m/2

Therefore, the slope of normal at (am^{2}, am^{3}) is

- 1/slope of the tangent at (am^{2}, am^{3})

= - 1/(3m / 2)

= - 2/3m

Hence, the equation of the normal at (am^{2}, am^{3}) is given by,

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

⇒ 3my - 3am^{4}

= - 2x + 2am^{2}

⇒ 2x +3my - am^{2} (2 + 3m^{2}) = 0

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

## Find the equation of the normal at the point (am^{2}, am^{3}) for the curve ay^{2} = x^{3}

**Summary:**

The equation of the normal at the point (am^{2}, am^{3}) for the curve ay^{2} = x^{3 }is 2x +3my - am^{2} (2 + 3m^{2}) = 0

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