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

Answer

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

### Step 1: Identify the equations of the curves
The given equations are:
1. Parabola: \( y^2 = 4x \)
2. Circle: \( (x - 6)^2 + y^2 = 1 \)

### Step 2: Determine the center and radius of the circle
From the equation of the circle, we can identify:
- Center \( C(6, 0) \)
- Radius \( r = 1 \)

### Step 3: Find the point on the parabola
Let the point on the parabola be \( P(t) \), where \( t \) is a parameter. The coordinates of point \( P \) on the parabola can be expressed as:
\[ P(t) = (t, 2\sqrt{t}) \]
since \( y^2 = 4x \) implies \( y = \pm 2\sqrt{x} \).

### Step 4: Calculate the slope of the tangent to the parabola
To find the slope of the tangent line to the parabola at point \( P(t) \), we differentiate \( y^2 = 4x \):
\[ \frac{d}{dx}(y^2) = \frac{d}{dx}(4x) \implies 2y \frac{dy}{dx} = 4 \implies \frac{dy}{dx} = \frac{2}{y} = \frac{2}{2\sqrt{t}} = \frac{1}{\sqrt{t}} \]
Thus, the slope of the tangent line at point \( P(t) \) is \( \frac{1}{\sqrt{t}} \).

### Step 5: Find the slope of the line connecting the center of the circle and point P
The slope \( m \) of the line connecting point \( C(6, 0) \) and point \( P(t) \) is given by:
\[ m = \frac{2\sqrt{t} - 0}{t - 6} = \frac{2\sqrt{t}}{t - 6} \]

### Step 6: Set up the perpendicularity condition
Since the shortest distance occurs along a line perpendicular to the tangent at point \( P(t) \), we have:
\[ \left( \frac{1}{\sqrt{t}} \right) \left( \frac{2\sqrt{t}}{t - 6} \right) = -1 \]
This leads to:
\[ \frac{2}{t - 6} = -1 \implies 2 = -t + 6 \implies t = 4 \]

### Step 7: Find the coordinates of point P
Substituting \( t = 4 \) into the coordinates of point \( P(t) \):
\[ P(4) = (4, 2\sqrt{4}) = (4, 4) \]

### Step 8: Calculate the distance from the center of the circle to point P
Now, we calculate the distance \( OM \) from the center \( C(6, 0) \) to point \( P(4, 4) \):
\[ OM = \sqrt{(6 - 4)^2 + (0 - 4)^2} = \sqrt{2^2 + (-4)^2} = \sqrt{4 + 16} = \sqrt{20} \]

### Step 9: Subtract the radius of the circle
The shortest distance \( d \) from the parabola to the circle is:
\[ d = OM - r = \sqrt{20} - 1 \]

### Final Answer
Thus, the shortest distance between the parabola and the circle is:
\[ d = \sqrt{20} - 1 \]