The point(s) on the parabola `y^2 = 4x` which are closest to the circle`x^2 + y^2 - 24y + 128 = 0` is/are

Options

A. `(0,0)`
B. `(2,2sqrt(2))`
C. `(4,4)`
D. none

Answer

To find the point(s) on the parabola \( y^2 = 4x \) that are closest to the circle given by the equation \( x^2 + y^2 - 24y + 128 = 0 \), we can follow these steps:

### Step 1: Rewrite the Circle's Equation
First, we need to rewrite the equation of the circle in standard form. The given equation is:
\[ x^2 + y^2 - 24y + 128 = 0 \]
We can complete the square for the \( y \) terms:
\[ x^2 + (y^2 - 24y) + 128 = 0 \]
\[ x^2 + (y - 12)^2 - 144 + 128 = 0 \]
\[ x^2 + (y - 12)^2 - 16 = 0 \]
\[ x^2 + (y - 12)^2 = 16 \]
This represents a circle with center \( (0, 12) \) and radius \( 4 \).

### Step 2: Parametrize the Parabola
The parabola \( y^2 = 4x \) can be parametrized as:
\[ x = t^2, \quad y = 2t \]
for some parameter \( t \).

### Step 3: Find the Distance from a Point on the Parabola to the Circle's Center
We need to find the distance from the point \( (t^2, 2t) \) on the parabola to the center of the circle \( (0, 12) \). The distance \( D \) is given by:
\[ D = \sqrt{(t^2 - 0)^2 + (2t - 12)^2} \]
Simplifying this, we have:
\[ D = \sqrt{t^4 + (2t - 12)^2} \]
\[ D = \sqrt{t^4 + (4t^2 - 48t + 144)} \]
\[ D = \sqrt{t^4 + 4t^2 - 48t + 144} \]

### Step 4: Minimize the Distance
To minimize \( D \), we can minimize \( D^2 \) (since the square root function is increasing):
\[ D^2 = t^4 + 4t^2 - 48t + 144 \]
Now, we differentiate \( D^2 \) with respect to \( t \) and set it to zero:
\[ \frac{d(D^2)}{dt} = 4t^3 + 8t - 48 \]
Setting the derivative to zero:
\[ 4t^3 + 8t - 48 = 0 \]
Dividing the entire equation by 4:
\[ t^3 + 2t - 12 = 0 \]

### Step 5: Solve the Cubic Equation
We can check for rational roots using the Rational Root Theorem. Testing \( t = 2 \):
\[ 2^3 + 2(2) - 12 = 8 + 4 - 12 = 0 \]
So, \( t = 2 \) is a root. We can factor the cubic polynomial:
\[ (t - 2)(t^2 + 2t + 6) = 0 \]
The quadratic \( t^2 + 2t + 6 \) has no real roots (discriminant \( 2^2 - 4 \cdot 1 \cdot 6 < 0 \)). Thus, the only real solution is \( t = 2 \).

### Step 6: Find the Corresponding Point on the Parabola
Substituting \( t = 2 \) back into the parametrization of the parabola:
\[ x = t^2 = 2^2 = 4, \quad y = 2t = 2(2) = 4 \]
Thus, the point on the parabola that is closest to the circle is \( (4, 4) \).

### Final Answer
The point on the parabola \( y^2 = 4x \) that is closest to the circle \( x^2 + y^2 - 24y + 128 = 0 \) is:
\[ \boxed{(4, 4)} \]