To find the point on the curve \( x^2 = 2y \) that is closest to the point \( (0, 5) \), we can follow these steps:
### Step 1: Define the points on the curve
The curve is given by the equation \( x^2 = 2y \). We can express points on the curve in terms of a variable \( h \) for \( x \) and \( k \) for \( y \):
\[
k = \frac{h^2}{2}
\]
So, any point on the curve can be represented as \( (h, \frac{h^2}{2}) \).
### Step 2: Write the distance formula
The distance \( D \) between the point \( (0, 5) \) and a point \( (h, k) \) on the curve is given by the distance formula:
\[
D = \sqrt{(h - 0)^2 + \left(\frac{h^2}{2} - 5\right)^2}
\]
This simplifies to:
\[
D = \sqrt{h^2 + \left(\frac{h^2}{2} - 5\right)^2}
\]
### Step 3: Simplify the distance expression
We can simplify the expression inside the square root:
\[
D = \sqrt{h^2 + \left(\frac{h^2}{2} - 5\right)^2}
\]
Expanding \( \left(\frac{h^2}{2} - 5\right)^2 \):
\[
\left(\frac{h^2}{2} - 5\right)^2 = \left(\frac{h^4}{4} - 5h^2 + 25\right)
\]
Thus, we have:
\[
D = \sqrt{h^2 + \frac{h^4}{4} - 5h^2 + 25} = \sqrt{\frac{h^4}{4} - 4h^2 + 25}
\]
### Step 4: Minimize the distance
To minimize \( D \), we can minimize \( D^2 \) (since the square root is a monotonically increasing function):
\[
D^2 = \frac{h^4}{4} - 4h^2 + 25
\]
Let \( f(h) = \frac{h^4}{4} - 4h^2 + 25 \).
### Step 5: Differentiate and find critical points
We differentiate \( f(h) \):
\[
f'(h) = h^3 - 8h
\]
Setting the derivative to zero to find critical points:
\[
h^3 - 8h = 0
\]
Factoring out \( h \):
\[
h(h^2 - 8) = 0
\]
This gives us:
\[
h = 0 \quad \text{or} \quad h^2 = 8 \quad \Rightarrow \quad h = \pm 2\sqrt{2}
\]
### Step 6: Find corresponding \( k \) values
For \( h = 0 \):
\[
k = \frac{0^2}{2} = 0 \quad \Rightarrow \quad (0, 0)
\]
For \( h = 2\sqrt{2} \):
\[
k = \frac{(2\sqrt{2})^2}{2} = \frac{8}{2} = 4 \quad \Rightarrow \quad (2\sqrt{2}, 4)
\]
For \( h = -2\sqrt{2} \):
\[
k = \frac{(-2\sqrt{2})^2}{2} = 4 \quad \Rightarrow \quad (-2\sqrt{2}, 4)
\]
### Step 7: Evaluate distances
Now we need to evaluate the distances from \( (0, 5) \) to both points \( (2\sqrt{2}, 4) \) and \( (-2\sqrt{2}, 4) \):
1. Distance to \( (2\sqrt{2}, 4) \):
\[
D = \sqrt{(2\sqrt{2} - 0)^2 + (4 - 5)^2} = \sqrt{(2\sqrt{2})^2 + (-1)^2} = \sqrt{8 + 1} = 3
\]
2. Distance to \( (-2\sqrt{2}, 4) \):
\[
D = \sqrt{(-2\sqrt{2} - 0)^2 + (4 - 5)^2} = \sqrt{(2\sqrt{2})^2 + (-1)^2} = \sqrt{8 + 1} = 3
\]
### Conclusion
Both points \( (2\sqrt{2}, 4) \) and \( (-2\sqrt{2}, 4) \) are equidistant from \( (0, 5) \) and are the closest points on the curve.
