Tangents are drawn to the parabola `y^2=4x` at the point P which is the upper end of latusrectum . Radius of the circle touching the parabola `y^2=4x` at the point P and passing through its focus is

Options

A. 1
B. `sqrt2`
C. `sqrt3`
D. 2

Answer

To solve the problem step by step, we need to find the radius of the circle that touches the parabola \(y^2 = 4x\) at the point \(P\) (the upper end of the latus rectum) and passes through its focus.

### Step 1: Identify the parabola and its properties
The given parabola is \(y^2 = 4x\).

- The focus of the parabola is at the point \((1, 0)\).
- The latus rectum of the parabola is a line segment perpendicular to the axis of symmetry of the parabola, passing through the focus. The endpoints of the latus rectum are at \((1, 2)\) and \((1, -2)\).

**Hint:** Remember that the focus of the parabola \(y^2 = 4px\) is at \((p, 0)\). Here, \(p = 1\).

### Step 2: Determine the point \(P\)
The upper end of the latus rectum \(P\) is at the point \((1, 2)\).

**Hint:** The coordinates of the upper end of the latus rectum can be found by substituting \(x = 1\) into the parabola's equation.

### Step 3: Write the equation of the tangent at point \(P\)
The equation of the tangent to the parabola \(y^2 = 4x\) at the point \(P(1, 2)\) can be found using the formula for the tangent line:

\[ yy_1 = 2p(x + x_1) \]

Substituting \(P(1, 2)\) where \(p = 1\):

\[ y \cdot 2 = 2 \cdot 1 (x + 1) \]

This simplifies to:

\[ 2y = 2x + 2 \quad \Rightarrow \quad y = x + 1 \]

**Hint:** The tangent line can be derived using the point-slope form of the tangent line at a given point on the parabola.

### Step 4: Set up the equation of the circle
The circle touches the parabola at point \(P(1, 2)\) and passes through the focus \((1, 0)\). The general equation of a circle can be written as:

\[ (x - h)^2 + (y - k)^2 = r^2 \]

Where \((h, k)\) is the center of the circle and \(r\) is the radius.

**Hint:** The center of the circle will be on the line perpendicular to the tangent at point \(P\).

### Step 5: Find the center of the circle
The slope of the tangent line \(y = x + 1\) is \(1\), so the slope of the radius (perpendicular to the tangent) is \(-1\). The equation of the line through \(P(1, 2)\) with slope \(-1\) is:

\[ y - 2 = -1(x - 1) \quad \Rightarrow \quad y = -x + 3 \]

Let the center of the circle be \((h, k)\). Since the circle passes through the focus \((1, 0)\):

\[ (1 - h)^2 + (0 - k)^2 = r^2 \]

**Hint:** Use the distance formula to find the radius in terms of the coordinates of the center.

### Step 6: Solve for the radius
Since the circle touches the parabola at \(P(1, 2)\), we also have:

\[ (1 - h)^2 + (2 - k)^2 = r^2 \]

Now we have two equations:

1. \((1 - h)^2 + (0 - k)^2 = r^2\)
2. \((1 - h)^2 + (2 - k)^2 = r^2\)

Setting these equal gives:

\[ (0 - k)^2 = (2 - k)^2 \]

Expanding this:

\[ k^2 = (2 - k)^2 \quad \Rightarrow \quad k^2 = 4 - 4k + k^2 \]

This simplifies to:

\[ 0 = 4 - 4k \quad \Rightarrow \quad k = 1 \]

Substituting \(k = 1\) back into either equation to find \(r\):

\[ (1 - h)^2 + (0 - 1)^2 = r^2 \quad \Rightarrow \quad (1 - h)^2 + 1 = r^2 \]

Using \(k = 1\) in the second equation:

\[ (1 - h)^2 + (2 - 1)^2 = r^2 \quad \Rightarrow \quad (1 - h)^2 + 1 = r^2 \]

Both equations are consistent.

### Step 7: Final calculation of the radius
The radius can be found from the distance from the center to the point \(P(1, 2)\):

\[ r = \sqrt{(1 - h)^2 + (2 - 1)^2} = \sqrt{(1 - h)^2 + 1} \]

Since we also have:

\[ r^2 = (1 - h)^2 + 1 \]

Thus, the radius \(r\) can be calculated as:

\[ r = \sqrt{2} \]

### Conclusion
The radius of the circle touching the parabola at point \(P\) and passing through its focus is:

\[ \boxed{\sqrt{2}} \]