Question

Find point on parabola $\ast$ $(x-1)^2 = -12(y-2)$ with focal distance 6.

• A. $(7, -1)$ and $(-5, -1)$
• B. $(1, 11)$ and $(1, -1)$
• C. $(7, 11)$ and $(-5, 11)$
• D. $(1, 2)$

Answer 1

Answer: (A)

The points on the parabola $(x-1)^2 = -12(y-2)$ with a focal distance of 6 are $(7, -1)$ and $(-5, -1)$.

Answer 2

1. Identify Parabola Parameters: The given equation is $(x-1)^2 = -12(y-2)$. This is in the standard form $(x-h)^2 = 4a(y-k)$, where $(h,k)$ is the vertex.

   • Vertex: $(h,k) = (1,2)$.
   • $4a = -12 \implies a = -3$.
   • Since $a$ is negative, the parabola opens downwards.

2. Determine Focus and Directrix:

   • Focus: $F = (h, k+a) = (1, 2 + (-3)) = (1, -1)$.
   • Directrix: $y = k-a = 2 - (-3) = 5$.

3. Use the Definition of Focal Distance: The focal distance of a point $P(x,y)$ on the parabola is the distance $PF$. We are given $PF = 6$. By the definition of a parabola, the distance from any point on the parabola to the focus is equal to its distance to the directrix. Let $PD$ be the distance from $P(x,y)$ to the directrix $y=5$. So, $PF = PD = |y-5|$. Given $PF = 6$, we have $|y-5| = 6$.

4. Solve for the y-coordinate: $|y-5| = 6$ implies two possibilities:

   • $y-5 = 6 \implies y = 11$.
   • $y-5 = -6 \implies y = -1$.

5. Check Validity of y-coordinates: The parabola opens downwards from its vertex $(1,2)$, so all points on the parabola must satisfy $y \le 2$.

   • $y=11$ violates $y \le 2$, so it's not a valid y-coordinate for a point on this parabola.
   • $y=-1$ satisfies $y \le 2$, so it's a valid y-coordinate.

6. Solve for the x-coordinate: Substitute $y=-1$ into the parabola's equation $(x-1)^2 = -12(y-2)$: $(x-1)^2 = -12(-1-2)$ $(x-1)^2 = -12(-3)$ $(x-1)^2 = 36$ Taking the square root of both sides: $x-1 = \pm 6$. This gives two possible x-values:

   • $x-1 = 6 \implies x = 7$.
   • $x-1 = -6 \implies x = -5$.

7. Identify the Points: The points on the parabola with focal distance 6 are $(7, -1)$ and $(-5, -1)$.