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 $ f(x) = \begin{cases} (1+ax)^{1/x} & , x<0 \\1+b & , x = 0 \\\frac{(x+4)^{1/2} - 2}{(x+c)^{1/3} - 2} & , x>0 \end{cases} $ be continuous at x = 0. Then $ e^a bc $ is equal to
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