Question

If the focus and vertex of a parabola are at a distance of 3 units and 6 units, respectively, from the origin on the positive x-axis, then the equation of the parabola is

• A. $y^{2}=-12(x-6)$
• B. $y^{2}=-9(x-6)$
• C. $y^{2}=-16(x-6)$
• D. $y=16(x-6)^{2}$
• E. $y^{2}=14(x-6)$

Hint:

Logic Tip: Always sketch a quick diagram for conic sections. Placing $V$ at $(6,0)$ and $S$ at $(3,0)$ visually proves the parabola opens leftwards, immediately indicating a negative coefficient for the linear $x$ term.

Answer 1

The Correct Option is A

Concept:
The standard equation of a horizontal parabola with vertex $(h, k)$ is $(y-k)^2 = 4a(x-h)$ if it opens to the right, and $(y-k)^2 = -4a(x-h)$ if it opens to the left. The value '$a$' represents the positive distance between the vertex and the focus.
Step 1: Identify the coordinates of the vertex and focus.
Both points lie on the positive x-axis ($y = 0$). Vertex $V$ is at a distance of 6 from the origin: $V(6, 0)$. Thus, $h = 6, k = 0$. Focus $S$ is at a distance of 3 from the origin: $S(3, 0)$.
Step 2: Determine the orientation and focal length (a).
The axis of symmetry is the x-axis. The focus $(3,0)$ is to the *left* of the vertex $(6,0)$. Because the focus is always "inside" the curve, the parabola must open to the left. This means we use the equation: $(y-k)^2 = -4a(x-h)$. The focal length $a$ is the distance from the vertex to the focus: $$a = 6 - 3 = 3$$
Step 3: Construct the equation.
Substitute $h = 6, k = 0$, and $a = 3$ into the standard equation: $$(y - 0)^2 = -4(3)(x - 6)$$ $$y^2 = -12(x - 6)$$