1. Core Concepts: Understanding Parabola Parameters and Focal Chord Properties

To solve this problem, we'll primarily use two key concepts:

• Parabola Standard Form and Focus: The equation of a parabola $y^2 = 4ax$ has its focus at $(a,0)$.
• Length of a Focal Chord: For a parabola $y^2 = 4ax$, if a focal chord makes an angle $\theta$ with the positive x-axis, its length $L$ is given by the formula: $L = 4a \csc^2 \theta$
• Distance of a Line from the Origin: The distance $p$ of a line $Ax + By + C = 0$ from the origin $(0,0)$ is given by: $p = \frac{|C|}{\sqrt{A^2 + B^2}}$

2. Analyzing the Given Parabola and its Focus

The given equation of the parabola is $y^2 = 12x$. Why this step? To work with any parabola problem, the first step is always to identify its standard parameters, especially the value of 'a', which defines the shape and position of the parabola and its focus.

We compare this to the standard form of a parabola $y^2 = 4ax$. By comparing the coefficients of $x$: $4a = 12$ $a = 3$

Now we can determine the focus of the parabola. Why this step? A focal chord, by definition, passes through the focus. Knowing the focus coordinates is essential for writing the equation of the focal chord. The focus $F$ of the parabola $y^2 = 4ax$ is at $(a,0)$. So, for our parabola, the focus $F$ is at $(3,0)$.

3. Utilizing the Length of the Focal Chord Formula

We are given that the length of the focal chord PQ is 15 units. Why this step? The length of the focal chord is a crucial piece of information. By using the formula for the length of a focal chord, we can determine the angle that the chord makes with the x-axis, which is necessary to define the chord's equation.

Using the formula $L = 4a \csc^2 \theta$: Substitute the given length $L = 15$ and the calculated value $a = 3$: $15 = 4(3) \csc^2 \theta$ $15 = 12 \csc^2 \theta$

Now, solve for $\csc^2 \theta$: $\csc^2 \theta = \frac{15}{12}$ $\csc^2 \theta = \frac{5}{4}$

From this, we can find $\sin^2 \theta$: $\sin^2 \theta = \frac{1}{\csc^2 \theta} = \frac{4}{5}$

And subsequently, $\cos^2 \theta$: $\cos^2 \theta = 1 - \sin^2 \theta = 1 - \frac{4}{5} = \frac{1}{5}$

Tip: Be careful with trigonometric identities. $\csc^2 \theta$ is $1/\sin^2 \theta$, not $\sin^2 \theta$.

4. Determining the Equation of the Focal Chord PQ

The focal chord PQ passes through the focus $F(3,0)$. Why this step? To find the distance of the chord from the origin, we first need to determine its equation. We have the angle information ($\sin^2 \theta$ and $\cos^2 \theta$) and a point it passes through (the focus).

The slope of the chord $m$ is given by $\tan \theta$. We can find $\tan^2 \theta$: $\tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta} = \frac{4/5}{1/5} = 4$ Therefore, the slope $m = \tan \theta = \pm \sqrt{4} = \pm 2$.

Now, we use the point-slope form of a linear equation, $y - y_1 = m(x - x_1)$, with $(x_1, y_1) = (3,0)$ and $m = \pm 2$. The equation of the focal chord PQ is: $y - 0 = (\pm 2)(x - 3)$ $y = \pm 2(x - 3)$

We can write this in the standard form $Ax + By + C = 0$: $\pm 2x - y \mp 6 = 0$ For calculation purposes, we can choose $m=2$ or $m=-2$. The final distance squared will be the same. Let's use $m=2$: $2x - y - 6 = 0$ If we chose $m=-2$: $-2x - y - (-6) = 0 \implies -2x - y + 6 = 0 \implies 2x + y - 6 = 0$. Notice that the constants are $\mp 6$.

Tip: The sign of the slope determines whether the chord goes "up-right" or "down-right" through the focus. However, the distance from the origin will be the same regardless of the sign of the slope, as it involves squaring the slope.

5. Calculating the Distance of the Chord from the Origin

We need to find the distance $p$ of the line $2x - y - 6 = 0$ (or $2x + y - 6 = 0$) from the origin $(0,0)$. Why this step? This is the direct application of the distance formula from the origin, which will give us the value of $p$ as required by the problem.

Using the distance formula $p = \frac{|C|}{\sqrt{A^2 + B^2}}$ for the line $Ax + By + C = 0$. For $2x - y - 6 = 0$: $A=2$, $B=-1$, $C=-6$. $p = \frac{|-6|}{\sqrt{(2)^2 + (-1)^2}}$ $p = \frac{6}{\sqrt{4 + 1}}$ $p = \frac{6}{\sqrt{5}}$

Tip: Always ensure the line equation is in the $Ax+By+C=0$ format before applying the distance formula. Also, remember the absolute value in the numerator.

6. Final Calculation: Finding $10p^2$

We have found the distance $p = \frac{6}{\sqrt{5}}$. Why this step? This is the final step to answer the specific question asked in the problem.

First, calculate $p^2$: $p^2 = \left(\frac{6}{\sqrt{5}}\right)^2 = \frac{6^2}{(\sqrt{5})^2} = \frac{36}{5}$

Finally, calculate $10p^2$: $10p^2 = 10 \times \frac{36}{5}$ $10p^2 = 2 \times 36$ $10p^2 = 72$

Wait, let me recheck the calculation. The question says the answer is 15. Let's recheck the problem statement and my steps.

Length of focal chord PQ = 15 units. $y^2 = 12x \implies a=3$. $L = 4a \csc^2 \theta$ $15 = 4(3) \csc^2 \theta$ $15 = 12 \csc^2 \theta$ $\csc^2 \theta = 15/12 = 5/4$. Correct. $\sin^2 \theta = 4/5$. Correct. $\cos^2 \theta = 1/5$. Correct. $\tan^2 \theta = \frac{4/5}{1/5} = 4$. Correct. $m = \tan \theta = \pm 2$. Correct.

Equation of focal chord through $(3,0)$ with slope $m$: $y - 0 = m(x - 3)$ $y = m(x - 3)$ $mx - y - 3m = 0$. Correct.

Distance from origin $p = \frac{|-3m|}{\sqrt{m^2 + (-1)^2}}$ $p = \frac{|-3(\pm 2)|}{\sqrt{(\pm 2)^2 + 1}}$ $p = \frac{|-6|}{\sqrt{4 + 1}}$ $p = \frac{6}{\sqrt{5}}$. Correct.

$p^2 = \frac{36}{5}$. Correct. $10p^2 = 10 \times \frac{36}{5} = 2 \times 36 = 72$.

My calculation leads to 72. The provided correct answer is 15. Let me check if there's an alternative interpretation or a common mistake I might have overlooked.

Is there another formula for the length of a focal chord? Yes, if the endpoints are $(at_1^2, 2at_1)$ and $(at_2^2, 2at_2)$, and it's a focal chord, then $t_1 t_2 = -1$. Length $L = a(t_1 - t_2)^2$. Also, $t_1 = \cot(\theta/2)$ and $t_2 = \tan(\theta/2)$. This is too complex. The formula $L = 4a \csc^2 \theta$ is standard and correct.

Let's re-read the problem very carefully. "Let the length of the focal chord PQ of the parabola $y^2=12 x$ be 15 units. If the distance of PQ from the origin is $p$, then $10 p^2$ is equal to _________."

Is it possible that the "origin" here refers to something else? No, "origin" typically means $(0,0)$.

Could the formula for the length of the focal chord be different? For $y^2 = 4ax$, the length of the focal chord making an angle $\theta$ with the x-axis is $4a \sec^2 \theta$ if the angle is with the axis of the parabola (x-axis in this case). No, it is $4a \csc^2 \theta$ if $\theta$ is the angle with the axis of the parabola. Let's derive it quickly to be sure. Parametric points on $y^2=4ax$ are $(at^2, 2at)$. If a chord passes through $(a,0)$, let the endpoints be $(at_1^2, 2at_1)$ and $(at_2^2, 2at_2)$. The slope of the line joining $(at_1^2, 2at_1)$ and $(a,0)$ is $m = \frac{2at_1 - 0}{at_1^2 - a} = \frac{2t_1}{t_1^2 - 1}$. The slope of the line joining $(at_2^2, 2at_2)$ and $(a,0)$ is $m = \frac{2at_2}{at_2^2 - a} = \frac{2t_2}{t_2^2 - 1}$. So $\frac{2t_1}{t_1^2 - 1} = \frac{2t_2}{t_2^2 - 1}$ implies $t_1(t