1. Understanding the Parabola's Equation and its Key Parameters

The foundation of any problem involving a conic section is correctly identifying its standard equation and associated parameters. A parabola is defined as the locus of a point that moves such that its distance from a fixed point (the focus) is equal to its distance from a fixed line (the directrix).

• Given Focus: $S(3,0)$
• Given Directrix: $x = -3$

For a standard parabola of the form $y^2 = 4ax$, the focus is at $(a,0)$ and the directrix is the line $x=-a$. This is a crucial standard form to recognize.

Why this step is important: By comparing the given information with the standard form, we can directly determine the value of 'a' for our specific parabola. This 'a' value is fundamental for all subsequent calculations, including the parametric representation of points and tangent properties.

Comparing the given focus $(3,0)$ with $(a,0)$, we find that $a=3$. Similarly, comparing the given directrix $x=-3$ with $x=-a$, we confirm $a=3$.

Now, substitute $a=3$ into the standard equation $y^2 = 4ax$: $y^2 = 4(3)x$ $y^2 = 12x$ This is the equation of the parabola we are working with.

Tip: Always begin by clearly identifying the parabola's equation and its 'a' value. A common mistake is to misidentify 'a' or to use the wrong standard form, which can lead to errors throughout the problem.

2. Parametric Representation of Points on the Parabola

When dealing with multiple points on a parabola, especially in problems involving ratios of coordinates or tangent intersections, using parametric coordinates is highly efficient.

Why use parametric form? For a parabola $y^2 = 4ax$, any point P on the parabola can be represented parametrically as $P(at^2, 2at)$, where $t$ is a parameter. This form simplifies calculations significantly because it reduces the number of variables (from $x, y$ to just $t$) and naturally incorporates the relationship between $x$ and $y$ coordinates on the parabola. It also makes tangent formulas much simpler.

From Step 1, we know that $a=3$ for our parabola $y^2 = 12x$. Let P and Q be the two points on the parabola. We can represent them using distinct parameters $t_1$ and $t_2$.

So, the coordinates of point P are: $P(3t_1^2, 2(3)t_1) \implies P(3t_1^2, 6t_1)$ And the coordinates of point Q are: $Q(3t_2^2, 2(3)t_2) \implies Q(3t_2^2, 6t_2)$

Here, $3t_1^2$ and $3t_2^2$ are the x-coordinates (abscissae), and $6t_1$ and $6t_2$ are the y-coordinates (ordinates).

3. Utilizing the Given Ratio of Ordinates

The problem states that the ordinates of points P and Q are in the ratio $3:1$. The ordinate refers to the y-coordinate of a point.

• Ordinate of P is $6t_1$.
• Ordinate of Q is $6t_2$.

Why this step is important: This piece of information provides a crucial relationship between the parameters $t_1$ and $t_2$. Establishing this relationship early simplifies subsequent calculations by allowing us to express one parameter in terms of the other.

According to the problem statement: $\frac{\text{Ordinate of P}}{\text{Ordinate of Q}} = \frac{3}{1}$ Substitute the parametric ordinates we found in Step 2: $\frac{6t_1}{6t_2} = \frac{3}{1}$ Simplify the equation by canceling the common factor of 6: $\frac{t_1}{t_2} = \frac{3}{1}$ This gives us the direct relationship: $t_1 = 3t_2$ This equation is vital for determining the coordinates of the intersection point of the tangents.

4. Finding the Intersection Point of Tangents R$(\alpha, \beta)$

The point of intersection of tangents drawn to a parabola at two distinct points is a standard result in coordinate geometry, significantly simplifying such problems.

Key Formula: If tangents are drawn to the parabola $y^2 = 4ax$ at points $P(at_1^2, 2at_1)$ and $Q(at_2^2, 2at_2)$, their point of intersection, $R(\alpha, \beta)$, is given by: $R(at_1t_2, a(t_1+t_2))$

Why this formula is used: This formula directly gives the coordinates of the intersection point, avoiding the lengthy process of finding the equations of individual tangents and then solving them simultaneously.

Now, we substitute the values we've determined:

• From Step 1, $a=3$.
• From Step 3, $t_1 = 3t_2$.

Let's find the x-coordinate, $\alpha$: $\alpha = a t_1 t_2$ Substitute $a=3$ and $t_1 = 3t_2$: $\alpha = 3 (3t_2) t_2$ $\alpha = 9t_2^2$

Next, let's find the y-coordinate, $\beta$: $\beta = a (t_1 + t_2)$ Substitute $a=3$ and $t_1 = 3t_2$: $\beta = 3 (3t_2 + t_2)$ $\beta = 3 (4t_2)$ $\beta = 12t_2$

So, the coordinates of the intersection point R are $(\alpha, \beta) = (9t_2^2, 12t_2)$.

Common Mistake: A frequent error is to forget the 'a' in the intersection formula, or to confuse the parameter 'a' with the 'a' in the point coordinates. Always write down the general formula first and then carefully substitute the values.

5. Calculating the Desired Expression $\frac{\beta^2}{\alpha}$

The final step is to compute the value of the expression $\frac{\beta^2}{\alpha}$ using the expressions for $\alpha$ and $\beta$ we just derived.

We have:

• $\alpha = 9t_2^2$
• $\beta = 12t_2$

Substitute these into the expression $\frac{\beta^2}{\alpha}$: $\frac{\beta^2}{\alpha} = \frac{(12t_2)^2}{9t_2^2}$

First, square the numerator: $\frac{\beta^2}{\alpha} = \frac{144t_2^2}{9t_2^2}$

Why we can simplify: Notice that $t_2^2$ appears in both the numerator and the denominator. Assuming $t_2 \neq 0$ (if $t_2=0$, then both P and Q would be at the vertex $(0,0)$, and their ordinates would be 0, which contradicts the ratio $3:1$), we can cancel out $t_2^2$: $\frac{\beta^2}{\alpha} = \frac{144}{9}$

Finally, perform the division: $\frac{144}{9} = 16$

Thus, the value of $\frac{\beta^2}{\alpha}$ is 16.

6. Summary and Key Takeaways

This problem effectively demonstrates the power of using parametric equations and standard formulas in coordinate geometry.

1. Parabola Identification: The first crucial step is always to correctly identify the parabola's equation ($y^2 = 12x$) and its parameter $a=3$ from the given focus and directrix.
2. Parametric Representation: Using the parametric form $P(at^2, 2at)$ for points on the parabola significantly simplifies handling coordinates and relationships.
3. Utilizing Given Conditions: The ratio of ordinates directly led to a simple relationship between the parameters $t_1 = 3t_2$. This is a common strategy in such problems.
4. Standard Tangent Formulas: Knowing the standard formula for the intersection of tangents $R(at_1t_2, a(t_1+t_2))$ saved a lot of time and potential calculation errors.
5. Algebraic Simplification: The final calculation involved straightforward substitution and simplification, where the parameter $t_2$ conveniently canceled out, yielding a numerical answer.

By following these systematic steps, even complex problems involving tangents and points on a parabola can be solved efficiently and accurately.