The equation of the normal to the curve `y^2=4ax` at point (a,2a) is

Options

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

Answer

To find the equation of the normal to the curve \( y^2 = 4ax \) at the point \( (a, 2a) \), we can follow these steps:

### Step 1: Differentiate the curve equation
We start with the curve equation:
\[ y^2 = 4ax \]
Differentiating both sides with respect to \( x \):
\[ 2y \frac{dy}{dx} = 4a \]
From this, we can solve for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{4a}{2y} = \frac{2a}{y} \]

### Step 2: Find the slope of the tangent at the given point
Now, we substitute the point \( (a, 2a) \) into the derivative to find the slope of the tangent:
\[ \frac{dy}{dx} \bigg|_{(a, 2a)} = \frac{2a}{2a} = 1 \]
Thus, the slope of the tangent line at the point \( (a, 2a) \) is \( 1 \).

### Step 3: Find the slope of the normal
The slope of the normal line is the negative reciprocal of the slope of the tangent. Therefore:
\[ m_n = -\frac{1}{m_t} = -\frac{1}{1} = -1 \]

### Step 4: Use the point-slope form to write the equation of the normal
Using the point-slope form of the equation of a line, which is given by:
\[ y - y_1 = m(x - x_1) \]
where \( (x_1, y_1) = (a, 2a) \) and \( m = -1 \):
\[ y - 2a = -1(x - a) \]
This simplifies to:
\[ y - 2a = -x + a \]

### Step 5: Rearrange the equation
Rearranging the equation gives:
\[ y + x - 3a = 0 \]
or equivalently:
\[ x + y - 3a = 0 \]

### Final Answer
Thus, the equation of the normal to the curve \( y^2 = 4ax \) at the point \( (a, 2a) \) is:
\[ \boxed{x + y - 3a = 0} \]

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