The circum circle of `triangle ABC` is `x^2 +y^2-5x-4y + 6=0` , if a parabola `(k + 1) y^2= x` have sides AB, BC, CA as tangents then the value of k is

Answer

To solve the problem, we need to find the value of \( k \) such that the parabola \( (k + 1)y^2 = x \) has its sides as tangents to the circumcircle of triangle \( ABC \) given by the equation \( x^2 + y^2 - 5x - 4y + 6 = 0 \).

### Step-by-Step Solution:

1. **Rewrite the Circle Equation**:
The circumcircle of triangle \( ABC \) is given by:
\[ x^2 + y^2 - 5x - 4y + 6 = 0 \]
We can complete the square for \( x \) and \( y \):
\[ (x^2 - 5x) + (y^2 - 4y) + 6 = 0 \]
Completing the square:
\[ (x - \frac{5}{2})^2 - \frac{25}{4} + (y - 2)^2 - 4 + 6 = 0 \]
Simplifying gives:
\[ (x - \frac{5}{2})^2 + (y - 2)^2 = \frac{25}{4} - 2 = \frac{17}{4} \]
Thus, the center of the circle is \( \left(\frac{5}{2}, 2\right) \) and the radius is \( \frac{\sqrt{17}}{2} \).

2. **Rewrite the Parabola Equation**:
The parabola is given by:
\[ (k + 1)y^2 = x \]
This can be rewritten as:
\[ y^2 = \frac{x}{k + 1} \]
This is a standard form of a parabola that opens to the right.

3. **Identify the Focus of the Parabola**:
The standard form \( y^2 = 4ax \) gives us \( 4a = \frac{1}{k + 1} \).
Therefore, the value of \( a \) is:
\[ a = \frac{1}{4(k + 1)} \]
The focus of the parabola is located at \( \left(a, 0\right) = \left(\frac{1}{4(k + 1)}, 0\right) \).

4. **Condition for Tangents**:
For the sides \( AB, BC, CA \) of triangle \( ABC \) to be tangents to the parabola, the circumcircle must pass through the focus of the parabola. Therefore, we substitute the focus into the circle equation:
\[ \left(\frac{1}{4(k + 1)} - \frac{5}{2}\right)^2 + (0 - 2)^2 = \frac{17}{4} \]

5. **Simplifying the Equation**:
Let \( p = \frac{1}{4(k + 1)} \). The equation becomes:
\[ \left(p - \frac{5}{2}\right)^2 + 4 = \frac{17}{4} \]
Simplifying gives:
\[ \left(p - \frac{5}{2}\right)^2 = \frac{17}{4} - 4 = \frac{17}{4} - \frac{16}{4} = \frac{1}{4} \]
Taking the square root:
\[ p - \frac{5}{2} = \frac{1}{2} \quad \text{or} \quad p - \frac{5}{2} = -\frac{1}{2} \]
Thus, we have two cases:
\[ p = 3 \quad \text{or} \quad p = 2 \]

6. **Finding \( k \)**:
Recall that \( p = \frac{1}{4(k + 1)} \):
- For \( p = 3 \):
\[ 3 = \frac{1}{4(k + 1)} \implies 4(k + 1) = \frac{1}{3} \implies k + 1 = \frac{1}{12} \implies k = \frac{1}{12} - 1 = -\frac{11}{12} \]
- For \( p = 2 \):
\[ 2 = \frac{1}{4(k + 1)} \implies 4(k + 1) = \frac{1}{2} \implies k + 1 = \frac{1}{8} \implies k = \frac{1}{8} - 1 = -\frac{7}{8} \]

### Final Values of \( k \):
Thus, the values of \( k \) are:
\[ k = -\frac{11}{12} \quad \text{and} \quad k = -\frac{7}{8} \]