Competency-Based
Assessment Made Simple.

The equation of the parabola is $y^2 = 12x$. Let the two points on the parabola be $P(x_1, y_1)$ and $Q(x_2, y_2)$. Since $P$ and $Q$ lie on the parabola, we have $y_1^2 = 12x_1$ and $y_2^2 = 12x_2$.

The chord joining $P$ and $Q$ intersects at right angles at the vertex $(0,0)$. The slope of the line joining $(x_1, y_1)$ and $(0,0)$ is $m_1 = \frac{y_1 - 0}{x_1 - 0} = \frac{y_1}{x_1}$. The slope of the line joining $(x_2, y_2)$ and $(0,0)$ is $m_2 = \frac{y_2 - 0}{x_2 - 0} = \frac{y_2}{x_2}$.

Since the lines intersect at right angles, the product of their slopes is $-1$. $m_1 m_2 = -1 \Rightarrow \frac{y_1}{x_1} \cdot \frac{y_2}{x_2} = -1 \Rightarrow y_1 y_2 = -x_1 x_2$.

We need to find the value of $x_1 x_2 - y_1 y_2$. Since $y_1^2 = 12x_1$ and $y_2^2 = 12x_2$, we have $x_1 = \frac{y_1^2}{12}$ and $x_2 = \frac{y_2^2}{12}$. Substituting these into the equation $y_1 y_2 = -x_1 x_2$, we get $y_1 y_2 = - \frac{y_1^2}{12} \cdot \frac{y_2^2}{12} \Rightarrow 1 = - \frac{y_1 y_2}{144} \Rightarrow y_1 y_2 = -144$.

Now, $x_1 x_2 = \frac{y_1^2}{12} \cdot \frac{y_2^2}{12} = \frac{(y_1 y_2)^2}{144} = \frac{(-144)^2}{144} = 144$.

Therefore, $x_1 x_2 - y_1 y_2 = 144 - (-144) = 144 + 144 = 288$.