Question

The length of the latus rectum of the parabola $y^2 = 12x$ and the focal distance of the point $(3, -6)$ is

• A. $3, 4$
• B. $2, 6$
• C. $-12, 6$
• D. $12, 6$

Hint:

The Focal Distance is simply the distance from the point to the focus. For $y^2=4ax$, the focus is $(a, 0)$. Using distance formula between $(3, -6)$ and $(3, 0)$: $\sqrt{(3-3)^2 + (-6-0)^2} = \sqrt{36} = 6$. It confirms our $x+a$ shortcut!

Answer 1

The Correct Option is D

Part 1: Length of Latus Rectum: Compare the given equation with the standard form $y^2 = 4ax$. $4a = 12 \implies a = 3$. The length of the latus rectum is defined as $4a$. $$\text{Length of Latus Rectum} = 12$$

Part 2: Focal Distance of point $(3, -6)$: The focal distance of any point $P(x_1, y_1)$ on the parabola $y^2 = 4ax$ is given by the formula: $$\text{Focal Distance} = x_1 + a$$ Here, the point is $(3, -6)$, so $x_1 = 3$. We already found $a = 3$. $$\text{Focal Distance} = 3 + 3 = 6$$ Thus, the required values are $12$ and $6$ respectively.