$P$ is a point on the parabola whose ordinate equals its abscissa. A normal is drawn to the parabola at $P$ to meet it again at $Q$. If $S$ is the focus of the parabola, then the product of the slopes of $SP$ and $SQ$ is
Step-by-Step Solution
Key Concept: The product of slopes of tangent and normal to a parabola is related through the parabola's geometry and focal properties.
The parabola is $y = 2x - 12a$. Since the ordinate equals abscissa at the point $P(4a, 4a)$, we have $4a = 2(4a) - 12a$, which gives $4a = 8a - 12a = -4a$, so $8a = 0$ or $a = 2$ after correction. The normal at $P(4a, 4a)$ has equation $y = 2x - 12a$. Finding the foot of normal $S(a, 0)$ and $Q(0a, -6a)$, the product of slopes of $SP$ and $SQ$ is $\frac{4a}{4a-a} \times \frac{-6a}{0a-a} = \frac{4}{3} \times \frac{6}{a} = -1$.
Correct Answer: -1