If two parabolas `y^(2)-4a(x-k)` and `x^(2)=4a(y-k)` have only one common point P, then the equation of normal to `y^(2)=4a(x-k)` at P is

Options

A. `y+x=4a`
B. `y+x=2a`
C. `y+x=4`
D. `y+x=2`

Answer

To solve the problem, we need to find the equation of the normal to the parabola \( y^2 = 4a(x - k) \) at the point \( P \) where it intersects with the other parabola \( x^2 = 4a(y - k) \). Given that these two parabolas have only one common point, we can follow these steps:

### Step 1: Identify the equations of the parabolas
The equations of the parabolas are:
1. \( y^2 = 4a(x - k) \) (Equation 1)
2. \( x^2 = 4a(y - k) \) (Equation 2)

### Step 2: Find the point of intersection
Since the two parabolas have only one common point, we can substitute \( y = x \) into either equation to find the coordinates of point \( P \).

Substituting \( y = x \) into Equation 1:
\[ x^2 = 4a(x - k) \]
This simplifies to:
\[ x^2 - 4ax + 4ak = 0 \]
For there to be only one solution (one common point), the discriminant of this quadratic equation must be zero:
\[ (-4a)^2 - 4 \cdot 1 \cdot 4ak = 0 \]
\[ 16a^2 - 16ak = 0 \]
Factoring out \( 16a \):
\[ 16a(a - k) = 0 \]
This gives us two cases: \( a = 0 \) (not valid for parabolas) or \( a = k \).

### Step 3: Find the coordinates of point \( P \)
If \( a = k \), substituting \( k \) back into the equation gives:
\[ x^2 = 4a(x - a) \]
This means:
\[ x^2 = 4a(x - a) \implies x^2 - 4ax + 4a^2 = 0 \]
The solution to this equation is:
\[ x = 2a \]
Thus, \( y = 2a \) as well. Therefore, the coordinates of point \( P \) are \( (2a, 2a) \).

### Step 4: Find the slope of the tangent at point \( P \)
To find the slope of the tangent line to the parabola \( y^2 = 4a(x - k) \) at point \( P \), we differentiate:
\[ 2y \frac{dy}{dx} = 4a \implies \frac{dy}{dx} = \frac{4a}{2y} = \frac{2a}{y} \]
At point \( P(2a, 2a) \):
\[ \frac{dy}{dx} = \frac{2a}{2a} = 1 \]
Thus, the slope of the tangent line at point \( P \) is \( 1 \).

### Step 5: Find the slope of the normal line
The slope of the normal line is the negative reciprocal of the slope of the tangent:
\[ \text{slope of normal} = -1 \]

### Step 6: Write the equation of the normal line
Using the point-slope form of the line equation:
\[ y - y_1 = m(x - x_1) \]
where \( (x_1, y_1) = (2a, 2a) \) and \( m = -1 \):
\[ y - 2a = -1(x - 2a) \]
Simplifying this:
\[ y - 2a = -x + 2a \implies y + x = 4a \]

### Final Answer
The equation of the normal to the parabola \( y^2 = 4a(x - k) \) at point \( P \) is:
\[ y + x = 4a \]