Question

The parabola with focus at \((4,-3)\) and vertex at \((4,-1)\) is

• A. \(x^2+8x+6y+22=0\)
• B. \(x^2-8x-10y+6=0\)
• C. \(x^2-8x-16y=0\)
• D. \(x^2-8x+8y+24=0\)

Hint:

For a vertical parabola, use \((x-h)^2=4a(y-k)\), where \((h,k)\) is the vertex and \((h,k+a)\) is the focus.

Answer 1

The Correct Option is D

Step 1: Identify the axis of the parabola.
Vertex $ = (4,-1) $ and focus $ = (4,-3) $ share the same $ x $-coordinate (both have $ x = 4 $), so the axis is vertical (parallel to the $ y $-axis).
Step 2: Determine the value of $ a $ (distance from vertex to focus).
In a vertical parabola $ (x-h)^2 = 4a(y-k) $, the focus is at $ (h, k+a) $. Given focus $ = (4,-3) $ and vertex $ = (4,-1) $: $ k + a = -3 $, so $ -1 + a = -3 $, giving $ a = -2 $ (negative means parabola opens downward).
Step 3: Write the standard-form equation.
\[ (x-4)^2 = 4(-2)(y+1) = -8(y+1) \]
Step 4: Expand to get general form.
\[ x^2 - 8x + 16 = -8y - 8 \] Rearranging: $ x^2 - 8x + 8y + 24 = 0 $.
Step 5: Verify by checking the vertex lies on the curve.
At $ (4,-1) $: $ 16 - 32 + 8(-1) + 24 = 16 - 32 - 8 + 24 = 0 $. Confirmed.
Step 6: State the answer.
\[ \boxed{x^2 - 8x + 8y + 24 = 0} \]