Let \( P \) and \( Q \) be points on the parabola \( y^2 = 12x \) with coordinates \( (x_1, y_1) \) and \( (x_2, y_2) \), respectively. Since \( (4, 1) \) is the midpoint of \( PQ \), we have:
\[ \frac{x_1 + x_2}{2} = 4 \implies x_1 + x_2 = 8, \] \[ \frac{y_1 + y_2}{2} = 1 \implies y_1 + y_2 = 2. \]
Since \( P \) and \( Q \) lie on the parabola \( y^2 = 12x \), we have:
\[ y_1^2 = 12x_1 \quad \text{and} \quad y_2^2 = 12x_2. \]
The equation of the chord of a parabola with a given midpoint can be derived as:
\[ y(y_1 + y_2) = 2x + x_1 + x_2. \]
Substituting \( y_1 + y_2 = 2 \) and \( x_1 + x_2 = 8 \), we get:
\[ y \cdot 2 = 2x + 8, \] \[ \implies y = x - 4. \]
Now, we substitute each option to check which one satisfies the equation \( y = x - 4 \).
Thus, the point \( \left(\frac{1}{2}, -20\right) \) lies on the line passing through points \( P \) and \( Q \).
Two parabolas have the same focus $(4, 3)$ and their directrices are the $x$-axis and the $y$-axis, respectively. If these parabolas intersect at the points $A$ and $B$, then $(AB)^2$ is equal to:
Let A be a 3 × 3 matrix such that \(\text{det}(A) = 5\). If \(\text{det}(3 \, \text{adj}(2A)) = 2^{\alpha \cdot 3^{\beta} \cdot 5^{\gamma}}\), then \( (\alpha + \beta + \gamma) \) is equal to: