The ordinates of the points $$P$$ and $$Q$$ on the parabola with focus $$(3, 0)$$ and directrix $$x = -3$$ are in the ratio 3 : 1. If $$R(\alpha, \beta)$$ is the point of intersection of the tangents to the parabola at $$P$$ and $$Q$$, then $$\frac{\beta^2}{\alpha}$$ is equal to _____.

To solve the parabola problem:

1. Identify the Parabola Equation

Given the Focus $$(3, 0)$$ and Directrix $$x = -3$$, the vertex is at the origin $$(0,0)$$. This is a standard right-opening parabola:

$$y^2 = 4ax \implies y^2 = 4(3)x \implies \mathbf{y^2 = 12x}$$

2. Find Point of Intersection $$R(\alpha, \beta)$$

Let the points on the parabola be $$P(at_1^2, 2at_1)$$ and $$Q(at_2^2, 2at_2)$$, where $$a = 3$$.

The intersection point $$R$$ of the tangents at $$P$$ and $$Q$$ is given by the formula:

$$R(\alpha, \beta) = (at_1t_2, a(t_1 + t_2))$$

3. Use the Ratio of Ordinates

The ordinates (y-coordinates) $$2at_1$$ and $$2at_2$$ are in the ratio $$3:1$$.

$$\frac{2at_1}{2at_2} = 3 \implies \mathbf{t_1 = 3t_2}$$

4. Express $$\alpha$$ and $$\beta$$ in terms of $$t_2$$

• $$\alpha = 3(3t_2)(t_2) = 9t_2^2$$
• $$\beta = 3(3t_2 + t_2) = 12t_2$$

5. Calculate $$\frac{\beta^2}{\alpha}$$

$$\frac{\beta^2}{\alpha} = \frac{(12t_2)^2}{9t_2^2} = \frac{144t_2^2}{9t_2^2} = \mathbf{16}$$

Final Answer: 16