Step 1: Understanding the Concept:
The shortest distance between two non-intersecting curves (or a point and a curve) always lies along their common normal. Therefore, the point $P$ on the parabola that is closest to the center of the circle must be a point where the normal to the parabola passes through the center of the circle.
Step 2: Key Formula or Approach:
1. Center of circle \(x^2 + y^2 + 2gx + 2fy + c = 0\) is \((-g, -f)\).
2. For \(x^2 = 4ay\), any point $P$ is \((2at, at^2)\). For \(x^2 = 4y\), $a=1$, so \(P = (2t, t^2)\).
3. Slope of tangent at $(2t, t^2)$ is $t$. Slope of normal is \(-1/t\).
Step 3: Detailed Explanation:
1. Identify Circle Center: Comparing \(x^2 + y^2 + 6x + 8 = 0\) with the standard form, \(2g = 6 \implies g = 3\). The center $C$ is \((-3, 0)\).
2. Normal at P: Point \(P = (2t, t^2)\). Differentiating \(x^2 = 4y \implies 2x = 4y' \implies y' = x/2\).
- Slope of tangent at $P = \frac{2t}{2} = t$.
- Slope of normal at $P = -\frac{1}{t}$.
3. Equate Normal Slopes: The line $CP$ is the normal.
- Slope $CP = \frac{t^2 - 0}{2t - (-3)} = \frac{t^2}{2t + 3}$.
- Set \(\frac{t^2}{2t + 3} = -\frac{1}{t} \implies t^3 = -2t - 3 \implies t^3 + 2t + 3 = 0\).
4. Solve for t: By inspection, $t = -1$ satisfies the equation (\(-1 - 2 + 3 = 0\)).
- Thus, point $P = (2(-1), (-1)^2) = (-2, 1)$.
5. Find Tangent Equation: Slope of tangent at $t = -1$ is $m = -1$.
- Equation: \(y - 1 = -1(x + 2) \implies y - 1 = -x - 2 \implies x + y + 1 = 0\).
Step 4: Final Answer
The equation of the tangent is \(x+y+1=0\).