If the line y-√3x+3=0 cuts the parabola y^2=x+2 at point P and Q then AP*AQ equals to, where A is(√3,0).

We are given the line equation:

y - √3x + 3 = 0
=> y = √3x - 3

And the parabola equation:

y² = x + 2

Step 1: Find Points of Intersection
Substituting y = √3x - 3 into the parabola equation:

(√3x - 3)² = x + 2
Expanding the square,

3x² - 6√3x + 9 = x + 2
Rearranging,

3x² - 6√3x + 9 - x - 2 = 0
=> 3x² - (6√3 + 1)x + 7 = 0

Step 2: Solve for x
The quadratic equation is:

3x² - (6√3 + 1)x + 7 = 0

Using the quadratic formula, x = (-b ± √(b² - 4ac)) / 2a,
where a = 3, b = -(6√3 + 1), and c = 7,

Discriminant:

Δ = b² - 4ac
= (6√3 + 1)² - 4(3)(7)
= (108 + 12√3 + 1) - 84
= 25 + 12√3

Thus, the roots are:

x = [(6√3 + 1) ± √(25 + 12√3)] / 6

Step 3: Find AP * AQ
The required value is given by the relation:

AP * AQ = c/a
= 7 / 3

Thus, the final answer is 7/3.