The equation of the line of the shortest distance between the parabola `y^(2)=4x` and the circle `x^(2)+y^(2)-4x-2y+4=0` is

Options

A. x + y = 3
B. x - y = 3
C. 2x + y = 5
D. None of these

Answer

To find the equation of the line of shortest distance between the parabola \( y^2 = 4x \) and the circle given by \( x^2 + y^2 - 4x - 2y + 4 = 0 \), we can follow these steps:

### Step 1: Identify the Parabola and Circle
The parabola is given by the equation:
\[ y^2 = 4x \]
This is a standard form of a parabola that opens to the right.

The circle is given by the equation:
\[ x^2 + y^2 - 4x - 2y + 4 = 0 \]
We can rewrite this in standard form.

### Step 2: Convert the Circle Equation to Standard Form
To convert the circle equation into standard form, we complete the square for both \( x \) and \( y \).

1. For \( x \):
\[ x^2 - 4x \rightarrow (x - 2)^2 - 4 \]
2. For \( y \):
\[ y^2 - 2y \rightarrow (y - 1)^2 - 1 \]

Substituting these back into the circle equation:
\[ (x - 2)^2 - 4 + (y - 1)^2 - 1 + 4 = 0 \]
This simplifies to:
\[ (x - 2)^2 + (y - 1)^2 - 1 = 0 \]
Thus, the standard form of the circle is:
\[ (x - 2)^2 + (y - 1)^2 = 1 \]
This shows that the center of the circle is \( (2, 1) \) and the radius is \( 1 \).

### Step 3: Find the Normal to the Parabola
The normal to the parabola \( y^2 = 4x \) at a point \( (at^2, 2at) \) is given by the equation:
\[ y - 2at = -\frac{1}{2a}(x - at^2) \]
For our parabola, \( a = 1 \), so the normal becomes:
\[ y - 2t = -\frac{1}{2}(x - t^2) \]
Rearranging gives:
\[ y = -\frac{1}{2}x + t + t^2 \]

### Step 4: Condition for the Normal to Pass Through the Circle's Center
Since the shortest distance line must also be a common normal, it must pass through the center of the circle \( (2, 1) \). Substituting \( x = 2 \) and \( y = 1 \) into the normal equation:
\[ 1 = -\frac{1}{2}(2) + t + t^2 \]
This simplifies to:
\[ 1 = -1 + t + t^2 \implies t + t^2 = 2 \]
Rearranging gives:
\[ t^2 + t - 2 = 0 \]

### Step 5: Solve the Quadratic Equation
We can solve the quadratic equation \( t^2 + t - 2 = 0 \) using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1 + 8}}{2} = \frac{-1 \pm 3}{2} \]
This gives us:
\[ t = 1 \quad \text{or} \quad t = -2 \]

### Step 6: Find the Equation of the Line
Substituting \( t = 1 \) into the normal equation:
\[ y = -\frac{1}{2}x + 1 + 1^2 = -\frac{1}{2}x + 2 \]
Rearranging gives:
\[ x + 2y = 4 \]

For \( t = -2 \):
\[ y = -\frac{1}{2}x - 2 + 4 = -\frac{1}{2}x + 2 \]
This gives the same line as before.

### Final Answer
The equation of the line of shortest distance is:
\[ x + y = 3 \]