For a parabola `y^(2)-2y-4x+9=0` the tangent at some point B is `3y=x+10`, where the normal at some other point K is `27y-9x+10=0`. Let `alpha` and `beta` are the segments of the chord BK cut by the axes of the parabola. Find the number of integral values of ‘a’ for which the equation `3x^(2)-(alpha+beta)x+(a^(2)-5a+ (353)/(27)) alpha beta=0` has its roots real and distinct

Explanation

To solve the given problem, we will follow these steps:

### Step 1: Rewrite the equation of the parabola
The equation of the parabola is given as:
\[ y^2 - 2y - 4x + 9 = 0 \]
We can rearrange this to express \( y \) in terms of \( x \):
\[ y^2 - 2y + 9 = 4x \]
This can be rewritten as:
\[ y^2 - 2y + 1 = 4x - 8 \]
\[ (y - 1)^2 = 4(x - 2) \]
This shows that the parabola opens to the right with vertex at \( (2, 1) \).

### Step 2: Find the point B where the tangent line touches the parabola
The tangent line at point B is given by:
\[ 3y = x + 10 \]
Rearranging gives:
\[ x - 3y + 10 = 0 \]
To find the point of tangency, we can substitute \( y \) from the tangent equation into the parabola's equation, but first, we need to express \( y \) in terms of \( x \):
\[ y = \frac{x + 10}{3} \]

Substituting \( y \) into the parabola's equation:
\[ \left(\frac{x + 10}{3}\right)^2 - 2\left(\frac{x + 10}{3}\right) - 4x + 9 = 0 \]
Expanding and simplifying gives us a quadratic equation in \( x \).

### Step 3: Find the point K where the normal line intersects the parabola
The normal line at point K is given by:
\[ 27y - 9x + 10 = 0 \]
Rearranging gives:
\[ 9x - 27y + 10 = 0 \]
To find the point of normality, we can again substitute \( y \) from this equation into the parabola's equation.

### Step 4: Determine the segments \( \alpha \) and \( \beta \)
The segments \( \alpha \) and \( \beta \) are the x-intercepts and y-intercepts of the chord BK. We can find these intercepts by substituting \( y = 0 \) and \( x = 0 \) into the equation of the line formed by points B and K.

### Step 5: Formulate the quadratic equation
We are given the quadratic equation:
\[ 3x^2 - (\alpha + \beta)x + \left(a^2 - 5a + \frac{353}{27}\right) \alpha \beta = 0 \]
For this equation to have real and distinct roots, the discriminant must be positive:
\[ D = (\alpha + \beta)^2 - 4 \cdot 3 \cdot \left(a^2 - 5a + \frac{353}{27}\right) \alpha \beta > 0 \]

### Step 6: Solve the inequality for \( a \)
We need to analyze the inequality derived from the discriminant condition to find the integral values of \( a \).