Find the point on `y=x^2+4` which is at shortest distance from the line `y=4x-4`

Answer

To find the point on the curve \( y = x^2 + 4 \) that is at the shortest distance from the line \( y = 4x - 4 \), we can follow these steps:

### Step 1: Understand the Problem
We need to find a point \((x_1, y_1)\) on the parabola \(y = x^2 + 4\) such that the distance to the line \(y = 4x - 4\) is minimized.

### Step 2: Set Up the Distance Formula
The distance \(d\) from a point \((x_1, y_1)\) to the line \(Ax + By + C = 0\) can be calculated using the formula:
\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
For the line \(y = 4x - 4\), we can rewrite it in the standard form:
\[ 4x - y - 4 = 0 \]
Here, \(A = 4\), \(B = -1\), and \(C = -4\).

### Step 3: Substitute the Point on the Parabola
Since \(y_1 = x_1^2 + 4\), we substitute \(y_1\) into the distance formula:
\[ d = \frac{|4x_1 - (x_1^2 + 4) - 4|}{\sqrt{4^2 + (-1)^2}} = \frac{|4x_1 - x_1^2 - 8|}{\sqrt{16 + 1}} = \frac{|4x_1 - x_1^2 - 8|}{\sqrt{17}} \]

### Step 4: Minimize the Distance
To minimize the distance \(d\), we need to minimize the expression inside the absolute value:
\[ f(x_1) = 4x_1 - x_1^2 - 8 \]
We will find the critical points by taking the derivative and setting it to zero:
\[ f'(x_1) = 4 - 2x_1 \]
Setting \(f'(x_1) = 0\):
\[ 4 - 2x_1 = 0 \implies 2x_1 = 4 \implies x_1 = 2 \]

### Step 5: Find the Corresponding \(y_1\)
Now, we substitute \(x_1 = 2\) back into the parabola equation to find \(y_1\):
\[ y_1 = (2)^2 + 4 = 4 + 4 = 8 \]

### Step 6: Conclusion
Thus, the point on the parabola \(y = x^2 + 4\) that is at the shortest distance from the line \(y = 4x - 4\) is:
\[ \boxed{(2, 8)} \]