This solution will guide you through the problem step-by-step, explaining the 'why' behind each action and leveraging key concepts from coordinate geometry.

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1. Understand the Parabola and its Focus

• Key Concept: The standard equation of a parabola opening to the right is $y^2 = 4ax$. Its focus is located at the point $S(a, 0)$. The parameter '$a$' defines the shape and position of the parabola.

• Given: We are given the equation of the parabola as $y^2 = 4x$.

• Step-by-step:

  1. Compare the given equation $y^2 = 4x$ with the standard form $y^2 = 4ax$.
  2. From this comparison, we can see that $4a = 4$.
  3. Solving for $a$, we get $a = 1$.
  4. Therefore, the focus of the parabola, denoted by $S$, is at $(a, 0) = (1, 0)$.

• Why this step is taken: Identifying the value of '$a$' and the coordinates of the focus $S$ is the foundational step. The problem involves a "focal chord," which, by definition, is a chord that passes through the focus. Knowing the focus is essential to define this chord.

• Tip: Always start by putting the parabola equation in its standard form to easily identify 'a', the focus, and the directrix.

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2. Determine the Coordinates of Point P on the Parabola

• Key Concept: Any point P on the parabola $y^2 = 4ax$ can be conveniently represented using parametric coordinates as $P(at^2, 2at)$, where $t$ is a parameter. This representation simplifies calculations involving points on the parabola.

• Applying the concept:

  1. Since we found $a=1$, the parametric coordinates of point P on the parabola $y^2=4x$ are $P(t^2, 2t)$.

• Key Concept: The slope of a line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the formula $m = \frac{y_2 - y_1}{x_2 - x_1}$.

• Applying the concept to the focal chord PS:

  1. The focal chord PS connects point $P(t^2, 2t)$ and the focus $S(1, 0)$.
  2. The slope of PS, denoted as $m_{PS}$, is $\frac{2t - 0}{t^2 - 1} = \frac{2t}{t^2 - 1}$.

• Key Concept: The slope of a line that makes an angle $\theta$ with the positive x-axis is given by $m = \tan \theta$.

• Given: The focal chord PQ (which is the same line as PS) makes an angle of $60^\circ$ with the positive x-axis.

• Step-by-step:

  1. The slope of the chord is $m_{PS} = \tan(60^\circ) = \sqrt{3}$.
  2. Equate the two expressions for the slope of PS: $\frac{2t}{t^2 - 1} = \sqrt{3}$
  3. Rearrange this equation to form a quadratic equation in $t$: $2t = \sqrt{3}(t^2 - 1)$ $\sqrt{3}t^2 - 2t - \sqrt{3} = 0$
  4. Solve this quadratic equation for $t$ using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: Here, $a=\sqrt{3}$, $b=-2$, $c=-\sqrt{3}$. $t = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(\sqrt{3})(-\sqrt{3})}}{2(\sqrt{3})}$ $t = \frac{2 \pm \sqrt{4 + 12}}{2\sqrt{3}}$ $t = \frac{2 \pm \sqrt{16}}{2\sqrt{3}}$ $t = \frac{2 \pm 4}{2\sqrt{3}}$
  5. This yields two possible values for $t$: $t_1 = \frac{2 + 4}{2\sqrt{3}} = \frac{6}{2\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}$ $t_2 = \frac{2 - 4}{2\sqrt{3}} = \frac{-2}{2\sqrt{3}} = -\frac{1}{\sqrt{3}}$

• Why this step is taken: We need to find the specific parameter 't' that defines point P. The given angle of the focal chord provides the necessary condition to determine 't'.

• Constraint: The problem states that point P lies in the first quadrant.

• Step-by-step:

  1. For $P(t^2, 2t)$ to be in the first quadrant, both its x-coordinate ($t^2$) and y-coordinate ($2t$) must be positive.
  2. $x = t^2 > 0$ is true for any $t \neq 0$.
  3. $y = 2t > 0$ implies $t > 0$.
  4. Comparing this condition with our two values for $t$, we must choose $t = \sqrt{3}$ (since $\sqrt{3} > 0$ and $-\frac{1}{\sqrt{3}} < 0$).

• Step-by-step:

  1. Substitute $t = \sqrt{3}$ back into the parametric coordinates of P: $P(t^2, 2t) = P((\sqrt{3})^2, 2\sqrt{3}) = P(3, 2\sqrt{3})$

• Tip: Always use given constraints (like quadrant information) to select the correct value from multiple possibilities. This prevents errors in subsequent calculations.

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**3. Equation