The equation of the normal to the curve `y^(4)=ax^(3)` at (a, a) is

Options

A. `x +2y=3a`
B. `3x-4y+a=0`
C. `4x+3y=7a`
D. `4x-3y=a`

Answer

To find the equation of the normal to the curve \( y^4 = ax^3 \) at the point \( (a, a) \), we can follow these steps:

### Step 1: Differentiate the curve
We start with the equation of the curve:
\[ y^4 = ax^3 \]
Differentiating both sides with respect to \( x \):
\[ \frac{d}{dx}(y^4) = \frac{d}{dx}(ax^3) \]
Using the chain rule on the left side, we get:
\[ 4y^3 \frac{dy}{dx} = 3ax^2 \]

### Step 2: Solve for \(\frac{dy}{dx}\)
Rearranging the equation to solve for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{3ax^2}{4y^3} \]

### 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. Thus:
\[ \text{slope of normal} = -\frac{1}{\frac{dy}{dx}} = -\frac{4y^3}{3ax^2} \]

### Step 4: Evaluate the slope at the point \((a, a)\)
Substituting \( x = a \) and \( y = a \) into the slope formula:
\[ \text{slope of normal at } (a, a) = -\frac{4a^3}{3a^2} = -\frac{4a}{3} \]

### Step 5: Write the equation of the normal
Using the point-slope form of the equation of a line:
\[ y - y_1 = m(x - x_1) \]
where \( (x_1, y_1) = (a, a) \) and \( m = -\frac{4a}{3} \):
\[ y - a = -\frac{4a}{3}(x - a) \]

### Step 6: Rearranging the equation
Expanding and rearranging:
\[ y - a = -\frac{4a}{3}x + \frac{4a^2}{3} \]
Multiplying through by 3 to eliminate the fraction:
\[ 3y - 3a = -4ax + 4a^2 \]
Rearranging gives:
\[ 4ax + 3y = 4a^2 + 3a \]

### Step 7: Final Equation
Thus, the equation of the normal is:
\[ 4x + 3y = 7a \]

### Final Answer
The equation of the normal to the curve \( y^4 = ax^3 \) at the point \( (a, a) \) is:
\[ 4x + 3y = 7a \]