The equation of the normal to the curve `y^2 = ax^3` at [a,a) is

Options

A. `x+2y=3a`
B. `x-4y=-a`
C. 4x+3y=7a
D. none of these

Answer

To find the equation of the normal to the curve \( y^2 = ax^3 \) at the point \( (a, a) \), we will follow these steps:

### Step 1: Differentiate the curve
We start with the equation of the curve:
\[ y^2 = ax^3 \]
To find the slope of the tangent line, we differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(y^2) = \frac{d}{dx}(ax^3) \]
Using the chain rule on the left side, we have:
\[ 2y \frac{dy}{dx} = 3ax^2 \]
Thus, we can express \( \frac{dy}{dx} \) as:
\[ \frac{dy}{dx} = \frac{3ax^2}{2y} \]

### Step 2: Evaluate the derivative at the point \( (a, a) \)
Next, we substitute \( x = a \) and \( y = a \) into the derivative:
\[ \frac{dy}{dx} \bigg|_{(a, a)} = \frac{3a \cdot a^2}{2a} = \frac{3a^3}{2a} = \frac{3a^2}{2} \]

### Step 3: Find the slope of the normal
The slope of the normal line is the negative reciprocal of the slope of the tangent line:
\[ m = -\frac{1}{\frac{dy}{dx}} = -\frac{2}{3a^2} \]

### Step 4: Use the point-slope form to write the equation of the normal
The equation of the normal line in point-slope form is given by:
\[ y - y_1 = m(x - x_1) \]
Substituting \( (x_1, y_1) = (a, a) \) and \( m = -\frac{2}{3a^2} \):
\[ y - a = -\frac{2}{3a^2}(x - a) \]

### Step 5: Rearranging the equation
Now, we will rearrange this equation:
\[ y - a = -\frac{2}{3a^2}x + \frac{2a}{3a^2} \]
Multiplying through by \( 3a^2 \) to eliminate the fraction:
\[ 3a^2(y - a) = -2x + 2a \]
Expanding and rearranging gives:
\[ 3a^2y - 3a^3 + 2x - 2a = 0 \]
Thus, we can write it as:
\[ 2x + 3a^2y - 3a^3 - 2a = 0 \]

### Final Equation
The equation of the normal to the curve \( y^2 = ax^3 \) at the point \( (a, a) \) is:
\[ 2x + 3a^2y = 3a^3 + 2a \]

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