P(-3, 2) is one end of focal chord PQ of the parabola `y^2+4x+4y=0`. Then the slope of the normal at Q is

Options

A. `-1//2`
B. `2`
C. `1//2`
D. `-2`

Answer

To solve the problem step by step, we will follow these instructions:

### Step 1: Rewrite the equation of the parabola
The given equation of the parabola is:
\[ y^2 + 4x + 4y = 0 \]
We can rearrange it to standard form:
\[ y^2 + 4y = -4x \]
Completing the square for \(y\):
\[ (y + 2)^2 - 4 = -4x \]
Thus, we have:
\[ (y + 2)^2 = -4(x + 1) \]
This represents a parabola that opens to the left with vertex at \((-1, -2)\).

### Step 2: Identify the coordinates of point P
The point \(P\) is given as \((-3, 2)\). We will use this point to find the slope of the tangent at \(P\).

### Step 3: Differentiate the equation to find the slope of the tangent
We differentiate the equation of the parabola implicitly:
\[ 2y \frac{dy}{dx} + 4 + 4\frac{dy}{dx} = 0 \]
Factoring out \(\frac{dy}{dx}\):
\[ (2y + 4)\frac{dy}{dx} + 4 = 0 \]
Solving for \(\frac{dy}{dx}\):
\[ (2y + 4)\frac{dy}{dx} = -4 \]
\[ \frac{dy}{dx} = \frac{-4}{2y + 4} \]

### Step 4: Substitute the coordinates of point P into the derivative
Now we substitute \(y = 2\) (the y-coordinate of point \(P\)):
\[ \frac{dy}{dx} = \frac{-4}{2(2) + 4} \]
\[ = \frac{-4}{4 + 4} = \frac{-4}{8} = -\frac{1}{2} \]
Thus, the slope of the tangent at point \(P\) is \(-\frac{1}{2}\).

### Step 5: Determine the slope of the normal at point Q
The slope of the normal is the negative reciprocal of the slope of the tangent. Therefore:
\[ \text{slope of normal} = -\frac{1}{\left(-\frac{1}{2}\right)} = 2 \]

### Final Answer
The slope of the normal at point \(Q\) is \(2\).

---