Area of Triangle Formed by Parabola Focus and Common Chord

To find the area of $\Delta PQS$, we first determine the coordinates of the focus $S$ and the intersection points $P$ and $Q$.

Step 1: Find the Focus $S$ of the Parabola

The given parabola is: $y^2 = 8x$

Comparing this with the standard equation of a parabola $y^2 = 4ax$: $4a = 8 \implies a = 2$

The coordinates of the focus $S$ are: $S = (a, 0) = (2, 0)$

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Step 2: Find the Intersection Points $P$ and $Q$

The equation of the circle is: $x^2 + y^2 - 2x - 4y = 0$

Substitute $y^2 = 8x$ into the equation of the circle: $x^2 + 8x - 2x - 4y = 0$ $x^2 + 6x - 4y = 0 \implies 4y = x^2 + 6x \implies y = \frac{x(x+6)}{4}$

Substitute this expression for $y$ back into $y^2 = 8x$: $\left(\frac{x(x+6)}{4}\right)^2 = 8x$ $\frac{x^2(x+6)^2}{16} = 8x$ $x\left[x(x+6)^2 - 128\right] = 0$

This yields one solution: $x = 0 \implies y = 0$ Thus, one intersection point is $P(0, 0)$.

For the remaining roots: $x(x^2 + 12x + 36) - 128 = 0$ $x^3 + 12x^2 + 36x - 128 = 0$

Testing $x = 2$: $2^3 + 12(2^2) + 36(2) - 128 = 8 + 48 + 72 - 128 = 0$

Factoring out $(x - 2)$: $(x - 2)(x^2 + 14x + 64) = 0$

The discriminant of the quadratic factor $x^2 + 14x + 64 = 0$ is: $\Delta = 14^2 - 4(1)(64) = 196 - 256 = -60 < 0$ Hence, there are no other real solutions.

For $x = 2$: $y = \frac{2(2+6)}{4} = \frac{16}{4} = 4$

Thus, the second intersection point is $Q(2, 4)$.

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Step 3: Calculate the Area of $\Delta PQS$

The vertices of $\Delta PQS$ are: $P(0, 0), \quad Q(2, 4), \quad S(2, 0)$

Notice that points $S(2, 0)$ and $Q(2, 4)$ both lie on the vertical line $x = 2$. Taking $SQ$ as the base: $\text{Base} = SQ = |4 - 0| = 4$

The perpendicular height from vertex $P(0, 0)$ to the line $x = 2$ is: $\text{Height} = |2 - 0| = 2$

Therefore, the area of $\Delta PQS$ is: $\text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times 4 \times 2 = 4$