Let the radius of the circle touching the parabola `y^(2)=x` at (1, 1) and having the directrix of `y^(2)=x` as its normal is equal to `ksqrt5` units, then k is equal to

Answer

To solve the problem, we need to find the value of \( k \) given that the radius of a circle touching the parabola \( y^2 = x \) at the point \( (1, 1) \) and having the directrix of the parabola as its normal is equal to \( k\sqrt{5} \) units.

### Step-by-Step Solution:

1. **Identify the Parabola and its Directrix**:
The equation of the parabola is \( y^2 = x \). This can be rewritten in the standard form \( y^2 = 4ax \) where \( 4a = 1 \). Thus, \( a = \frac{1}{4} \).
The directrix of the parabola \( y^2 = 4ax \) is given by \( x = -a \). Therefore, the directrix is:
\[ x = -\frac{1}{4} \]

2. **Find the Normal at the Point (1, 1)**:
The slope of the tangent to the parabola at any point can be found using implicit differentiation. Differentiating \( y^2 = x \) gives:
\[ 2y \frac{dy}{dx} = 1 \quad \Rightarrow \quad \frac{dy}{dx} = \frac{1}{2y} \]
At the point \( (1, 1) \):
\[ \frac{dy}{dx} = \frac{1}{2 \cdot 1} = \frac{1}{2} \]
The slope of the normal is the negative reciprocal of the slope of the tangent:
\[ \text{slope of normal} = -\frac{1}{\frac{1}{2}} = -2 \]

3. **Equation of the Normal**:
Using the point-slope form of the equation of a line, the equation of the normal at \( (1, 1) \) is:
\[ y - 1 = -2(x - 1) \quad \Rightarrow \quad y = -2x + 3 \]

4. **Find the Center of the Circle**:
Let the center of the circle be \( C(-\frac{1}{4}, a) \). Since the normal line passes through the center \( C \), we can substitute \( x = -\frac{1}{4} \) into the normal's equation to find \( a \):
\[ a = -2\left(-\frac{1}{4}\right) + 3 = \frac{1}{2} + 3 = \frac{7}{2} \]
Thus, the coordinates of the center \( C \) are \( \left(-\frac{1}{4}, \frac{7}{2}\right) \).

5. **Calculate the Radius of the Circle**:
The radius \( r \) of the circle is the distance from the center \( C \) to the point of tangency \( P(1, 1) \):
\[ r = \sqrt{\left(1 - \left(-\frac{1}{4}\right)\right)^2 + \left(1 - \frac{7}{2}\right)^2} \]
Simplifying the x-coordinate:
\[ 1 - \left(-\frac{1}{4}\right) = 1 + \frac{1}{4} = \frac{5}{4} \]
And simplifying the y-coordinate:
\[ 1 - \frac{7}{2} = \frac{2}{2} - \frac{7}{2} = -\frac{5}{2} \]
Now substituting these values into the radius formula:
\[ r = \sqrt{\left(\frac{5}{4}\right)^2 + \left(-\frac{5}{2}\right)^2} = \sqrt{\frac{25}{16} + \frac{25}{4}} = \sqrt{\frac{25}{16} + \frac{100}{16}} = \sqrt{\frac{125}{16}} = \frac{5\sqrt{5}}{4} \]

6. **Relate Radius to \( k\sqrt{5} \)**:
According to the problem, the radius is given as \( k\sqrt{5} \):
\[ \frac{5\sqrt{5}}{4} = k\sqrt{5} \]
Dividing both sides by \( \sqrt{5} \):
\[ \frac{5}{4} = k \]

### Final Answer:
Thus, the value of \( k \) is:
\[ \boxed{\frac{5}{4}} \]