Let P be a point on parabola ` x ^ 2 = 4y ` . If the distance of P from the centre of circle ` x ^ 2 + y ^ 2 + 6x + 8 = 0 ` is minimum, then the equation of tangent at P on parabola ` x ^ 2 = 4y ` is :

Options

A. ` x + y + 1 = 0 `
B. ` x + y - 1 = 0`
C. ` x - y + 1 = 0`
D. ` x - y - 1 = 0`

Explanation

To solve the problem, we need to find the equation of the tangent to the parabola \( x^2 = 4y \) at the point \( P \) such that the distance from \( P \) to the center of the circle given by \( x^2 + y^2 + 6x + 8 = 0 \) is minimized.

### Step 1: Find the center and radius of the circle
The equation of the circle is given as:
\[ x^2 + y^2 + 6x + 8 = 0 \]
To find the center, we can rewrite this in standard form. We complete the square for the \( x \) terms:
\[ (x^2 + 6x) + y^2 + 8 = 0 \]
Completing the square for \( x^2 + 6x \):
\[ (x + 3)^2 - 9 + y^2 + 8 = 0 \]
\[ (x + 3)^2 + y^2 - 1 = 0 \]
This gives us the center of the circle at \( (-3, 0) \) and the radius as \( r = 1 \).

### Step 2: Parameterize the point \( P \) on the parabola
The parabola \( x^2 = 4y \) can be parameterized as:
\[ P(x, y) = (x, \frac{x^2}{4}) \]

### Step 3: Find the distance from point \( P \) to the center of the circle
The distance \( d \) from point \( P \) to the center \( C(-3, 0) \) is given by:
\[ d = \sqrt{(x + 3)^2 + \left(\frac{x^2}{4}\right)^2} \]
Squaring the distance (to simplify calculations):
\[ d^2 = (x + 3)^2 + \left(\frac{x^2}{4}\right)^2 \]
\[ = (x + 3)^2 + \frac{x^4}{16} \]

### Step 4: Minimize the distance squared
To minimize \( d^2 \), we take the derivative with respect to \( x \) and set it to zero:
\[ \frac{d}{dx}\left((x + 3)^2 + \frac{x^4}{16}\right) = 2(x + 3) + \frac{4x^3}{16} = 2(x + 3) + \frac{x^3}{4} \]
Setting the derivative equal to zero:
\[ 2(x + 3) + \frac{x^3}{4} = 0 \]
Multiplying through by 4 to eliminate the fraction:
\[ 8(x + 3) + x^3 = 0 \]
\[ x^3 + 8x + 24 = 0 \]

### Step 5: Solve the cubic equation
We can try \( x = -2 \):
\[ (-2)^3 + 8(-2) + 24 = -8 - 16 + 24 = 0 \]
Thus, \( x = -2 \) is a root. We can factor the cubic:
\[ x^3 + 8x + 24 = (x + 2)(x^2 - 2x + 12) \]
The quadratic \( x^2 - 2x + 12 \) has no real roots (discriminant \( < 0 \)). Therefore, the only real solution is \( x = -2 \).

### Step 6: Find the corresponding \( y \)
Substituting \( x = -2 \) into the parabola equation:
\[ y = \frac{(-2)^2}{4} = 1 \]
Thus, the point \( P \) is \( (-2, 1) \).

### Step 7: Find the slope of the tangent line
The slope of the tangent line at point \( P \) on the parabola \( x^2 = 4y \) is given by:
\[ \frac{dy}{dx} = \frac{1}{2}x \]
At \( x = -2 \):
\[ \frac{dy}{dx} = \frac{1}{2}(-2) = -1 \]

### Step 8: Write the equation of the tangent line
Using the point-slope form of the line:
\[ y - y_1 = m(x - x_1) \]
Substituting \( P(-2, 1) \) and \( m = -1 \):
\[ y - 1 = -1(x + 2) \]
\[ y - 1 = -x - 2 \]
\[ y = -x - 1 \]

### Final Answer
The equation of the tangent at point \( P \) on the parabola is:
\[ \boxed{y + x + 1 = 0} \]