Area of Quadrilateral Formed by Points of Tangency

To find the area of the quadrilateral $PQRS$, we first determine the equations of the common tangents to the circle and the parabola.

Step 1: Equation of Tangents to the Parabola

The equation of the parabola is: $y^2 = 8x$ Comparing this with the standard form $y^2 = 4ax$, we have $a = 2$.

The equation of a tangent to the parabola $y^2 = 4ax$ with slope $m$ is: $y = mx + \frac{a}{m} \implies y = mx + \frac{2}{m}$

Step 2: Condition for Tangency to the Circle

The given circle is: $x^2 + y^2 = 2$ The circle is centered at $(0, 0)$ with radius $r = \sqrt{2}$.

For the line $mx - y + \frac{2}{m} = 0$ to be tangent to the circle, the perpendicular distance from the center $(0,0)$ to the line must be equal to the radius $r$: $\left| \frac{\frac{2}{m}}{\sqrt{1 + m^2}} \right| = \sqrt{2}$

Squaring both sides: $\frac{4}{m^2(1 + m^2)} = 2 \implies m^2(1 + m^2) = 2$

Letting $u = m^2$ ($u > 0$): $u^2 + u - 2 = 0 \implies (u + 2)(u - 1) = 0$

Since $u > 0$, we have $u = 1$, which gives: $m = \pm 1$

Thus, the two common tangents are:

1. For $m = 1$: $y = x + 2 \implies x - y + 2 = 0$
2. For $m = -1$: $y = -x - 2 \implies x + y + 2 = 0$

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Step 3: Finding Points of Contact on the Parabola ($R$ and $S$)

The point of tangency for the tangent line $y = mx + \frac{a}{m}$ on the parabola $y^2 = 4ax$ is given by: $\left( \frac{a}{m^2}, \frac{2a}{m} \right)$

Substituting $a = 2$:

• For $m = 1$: $R = \left( \frac{2}{1^2}, \frac{2(2)}{1} \right) = (2, 4)$
• For $m = -1$: $S = \left( \frac{2}{(-1)^2}, \frac{2(2)}{-1} \right) = (2, -4)$

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Step 4: Finding Points of Contact on the Circle ($P$ and $Q$)

The equation of the tangent at $(x_1, y_1)$ on the circle $x^2 + y^2 = 2$ is: $x x_1 + y y_1 = 2$

• Comparing $x x_1 + y y_1 = 2$ with $-x + y = 2$ (for $m = 1$): $P = (-1, 1)$
• Comparing $x x_1 + y y_1 = 2$ with $-x - y = 2$ (for $m = -1$): $Q = (-1, -1)$

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Step 5: Area of Quadrilateral $PQRS$

The vertices of the quadrilateral are: $P(-1, 1), \quad Q(-1, -1), \quad R(2, 4), \quad S(2, -4)$

• The line segment $PQ$ lies on the vertical line $x = -1$, with length: $PQ = |1 - (-1)| = 2$
• The line segment $RS$ lies on the vertical line $x = 2$, with length: $RS = |4 - (-4)| = 8$

Since both $PQ$ and $RS$ are vertical, they are parallel to each other ($PQ \parallel RS$). Thus, $PQRS$ is a trapezoid.

The perpendicular distance (height $h$) between the two parallel sides is: $h = 2 - (-1) = 3$

The area of the trapezoid is given by: $\text{Area} = \frac{1}{2} (PQ + RS) \times h = \frac{1}{2} (2 + 8) \times 3 = \frac{1}{2} \times 10 \times 3 = 15$

Therefore, the correct option is (D).