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Find the equation of the normal to the curve `ay^(2)=x^(3)` at the point `(am^(2), am^(3))`.

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To find the equation of the normal to the curve \( ay^2 = x^3 \) at the point \( (am^2, am^3) \), we will follow these steps: ### Step 1: Differentiate the curve We start with the equation of the curve: \[ ay^2 = x^3 \] To find the slope of the tangent, we differentiate both sides with respect to \( x \): \[ \frac{d}{dx}(ay^2) = \frac{d}{dx}(x^3) \] Using the chain rule on the left side, we get: \[ 2ay \frac{dy}{dx} = 3x^2 \] ### Step 2: Solve for \(\frac{dy}{dx}\) Rearranging the equation to solve for \(\frac{dy}{dx}\): \[ \frac{dy}{dx} = \frac{3x^2}{2ay} \] ### Step 3: Substitute the point into the derivative Now, we substitute the point \( (am^2, am^3) \) into the derivative: \[ x = am^2 \quad \text{and} \quad y = am^3 \] Substituting these values into the derivative: \[ \frac{dy}{dx} = \frac{3(am^2)^2}{2a(am^3)} = \frac{3a^2m^4}{2a^2m^3} = \frac{3m}{2} \] ### Step 4: Find the slope of the normal The slope of the tangent line at the point is \( \frac{3m}{2} \). The slope of the normal line is the negative reciprocal of the slope of the tangent: \[ \text{slope of normal} = -\frac{1}{\frac{3m}{2}} = -\frac{2}{3m} \] ### Step 5: Use point-slope form to find the equation of the normal Using the point-slope form of the equation of a line: \[ y - y_1 = m(x - x_1) \] where \( m \) is the slope of the normal, and \( (x_1, y_1) = (am^2, am^3) \): \[ y - am^3 = -\frac{2}{3m}(x - am^2) \] ### Step 6: Simplify the equation Expanding this equation: \[ y - am^3 = -\frac{2}{3m}x + \frac{2}{3m}(am^2) \] Rearranging gives: \[ y = -\frac{2}{3m}x + \frac{2a m}{3} + am^3 \] ### Step 7: Combine terms To combine the constant terms: \[ y = -\frac{2}{3m}x + \left(\frac{2a m + 3a m^3}{3}\right) \] Thus, the equation of the normal line can be written as: \[ 2x + 3my - (2am + 3am^3) = 0 \] ### Final Answer The equation of the normal to the curve \( ay^2 = x^3 \) at the point \( (am^2, am^3) \) is: \[ 2x + 3my - (2am + 3am^3) = 0 \]
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