Find shortest distance between `y^(2) =4x" and "(x-6)^(2) +y^(2) =1`

Answer

To find the shortest distance between the curves \( y^2 = 4x \) (a parabola) and \( (x-6)^2 + y^2 = 1 \) (a circle), we can follow these steps:

### Step 1: Understand the Curves
The first curve \( y^2 = 4x \) is a parabola opening to the right. The second curve \( (x-6)^2 + y^2 = 1 \) is a circle with center at \( (6, 0) \) and radius \( 1 \).

### Step 2: Parametrize the Parabola
Let’s parametrize the parabola. We can set \( x = t \), then from the equation of the parabola:
\[ y^2 = 4t \implies y = 2\sqrt{t} \]
Thus, a point on the parabola can be represented as \( P(t, 2\sqrt{t}) \).

### Step 3: Find the Slope of the Tangent to the Parabola
Differentiate \( y^2 = 4x \) with respect to \( x \):
\[ 2y \frac{dy}{dx} = 4 \implies \frac{dy}{dx} = \frac{4}{2y} = \frac{2}{\sqrt{t}} \]
The slope of the tangent line at point \( P(t, 2\sqrt{t}) \) is \( \frac{2}{\sqrt{t}} \).

### Step 4: Find the Center of the Circle
The center of the circle \( (x-6)^2 + y^2 = 1 \) is at \( C(6, 0) \) and the radius is \( 1 \).

### Step 5: Find the Slope of the Line from Point on Parabola to Center of Circle
The slope of the line \( PC \) from point \( P(t, 2\sqrt{t}) \) to center \( C(6, 0) \) is given by:
\[ \text{slope of } PC = \frac{0 - 2\sqrt{t}}{6 - t} = \frac{-2\sqrt{t}}{6 - t} \]

### Step 6: Set Up the Condition for Perpendicularity
Since the tangent at point \( P \) and the line \( PC \) are perpendicular, their slopes must satisfy:
\[ \frac{2}{\sqrt{t}} \cdot \frac{-2\sqrt{t}}{6 - t} = -1 \]
This simplifies to:
\[ \frac{-4}{6 - t} = -1 \implies 4 = 6 - t \implies t = 2 \]

### Step 7: Find the Coordinates of Point P
Substituting \( t = 2 \) back into the parametrization:
\[ P(2, 2\sqrt{2}) = (2, 2\sqrt{2}) \]

### Step 8: Calculate the Distance from Point P to Center C
Now, we find the distance \( PC \):
\[ PC = \sqrt{(6 - 2)^2 + (0 - 2\sqrt{2})^2} = \sqrt{4 + (2\sqrt{2})^2} = \sqrt{4 + 8} = \sqrt{12} = 2\sqrt{3} \]

### Step 9: Subtract the Radius of the Circle
The shortest distance \( d \) from the parabola to the circle is:
\[ d = PC - \text{radius} = 2\sqrt{3} - 1 \]

### Final Answer
Thus, the shortest distance between the curves \( y^2 = 4x \) and \( (x-6)^2 + y^2 = 1 \) is:
\[ \boxed{2\sqrt{3} - 1} \]