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=0`

Explanation

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
First, we need to find the derivative \( \frac{dy}{dx} \) of the given curve. We start with the equation:

\[ y^4 = ax^3 \]

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 \( \frac{dy}{dx} \)
Now, we can isolate \( \frac{dy}{dx} \):

\[ \frac{dy}{dx} = \frac{3ax^2}{4y^3} \]

### Step 3: Find the slope of the tangent at the point \( (a, a) \)
Next, we substitute \( x = a \) and \( y = a \) into the derivative to find the slope of the tangent line at the point \( (a, a) \):

\[ \frac{dy}{dx} \bigg|_{(a, a)} = \frac{3a(a^2)}{4a^3} = \frac{3a^3}{4a^3} = \frac{3}{4} \]

### Step 4: Find the slope of the normal
The slope of the normal line is the negative reciprocal of the slope of the tangent line. Therefore, the slope \( m \) of the normal line is:

\[ m = -\frac{1}{\frac{3}{4}} = -\frac{4}{3} \]

### Step 5: Use the point-slope form to find the equation of the normal
We can use the point-slope form of the equation of a line, which is given by:

\[ y - y_1 = m(x - x_1) \]

Substituting \( m = -\frac{4}{3} \) and the point \( (a, a) \):

\[ y - a = -\frac{4}{3}(x - a) \]

### Step 6: Rearranging the equation
Now, we rearrange the equation to get it into a standard form:

\[ y - a = -\frac{4}{3}x + \frac{4}{3}a \]

Adding \( a \) to both sides gives:

\[ y = -\frac{4}{3}x + \frac{4}{3}a + a \]

Combining the terms:

\[ y = -\frac{4}{3}x + \frac{4a}{3} + \frac{3a}{3} = -\frac{4}{3}x + \frac{7a}{3} \]

### Step 7: Convert to standard form
To convert this into standard form, we can multiply through by 3 to eliminate the fraction:

\[ 3y = -4x + 7a \]

Rearranging gives:

\[ 4x + 3y = 7a \]

Thus, the equation of the normal to the curve at the point \( (a, a) \) is:

\[ \boxed{4x + 3y = 7a} \]