Question:
The parabola with focus at \((4,-3)\) and vertex at \((4,-1)\) is
The parabola with focus at \((4,-3)\) and vertex at \((4,-1)\) is
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For a vertical parabola, use \((x-h)^2=4a(y-k)\), where \((h,k)\) is the vertex and \((h,k+a)\) is the focus.
Updated On: Jun 25, 2026
- \(x^2+8x+6y+22=0\)
- \(x^2-8x-10y+6=0\)
- \(x^2-8x-16y=0\)
- \(x^2-8x+8y+24=0\)
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The Correct Option is D
Solution and Explanation
Step 1: Identify the vertex and focus.
Vertex is \[ (4,-1) \] Focus is \[ (4,-3) \] Since both points have the same \(x\)-coordinate, the axis of the parabola is vertical.
Step 2: Find the value of \(a\).
For a vertical parabola with vertex \((h,k)\), standard form is \[ (x-h)^2=4a(y-k) \] Here, \[ (h,k)=(4,-1) \] Focus is \[ (h,k+a) \] So, \[ (4,-1+a)=(4,-3) \] Therefore, \[ -1+a=-3 \] \[ a=-2 \]
Step 3: Write the equation of the parabola.
Using \[ (x-h)^2=4a(y-k), \] we get \[ (x-4)^2=4(-2)(y+1) \] \[ (x-4)^2=-8(y+1) \]
Step 4: Convert into general form.
Expanding, \[ x^2-8x+16=-8y-8 \] Bring all terms to one side: \[ x^2-8x+8y+24=0 \]
Step 5: Final conclusion.
Therefore, the required parabola is \[ \boxed{x^2-8x+8y+24=0} \]
Vertex is \[ (4,-1) \] Focus is \[ (4,-3) \] Since both points have the same \(x\)-coordinate, the axis of the parabola is vertical.
Step 2: Find the value of \(a\).
For a vertical parabola with vertex \((h,k)\), standard form is \[ (x-h)^2=4a(y-k) \] Here, \[ (h,k)=(4,-1) \] Focus is \[ (h,k+a) \] So, \[ (4,-1+a)=(4,-3) \] Therefore, \[ -1+a=-3 \] \[ a=-2 \]
Step 3: Write the equation of the parabola.
Using \[ (x-h)^2=4a(y-k), \] we get \[ (x-4)^2=4(-2)(y+1) \] \[ (x-4)^2=-8(y+1) \]
Step 4: Convert into general form.
Expanding, \[ x^2-8x+16=-8y-8 \] Bring all terms to one side: \[ x^2-8x+8y+24=0 \]
Step 5: Final conclusion.
Therefore, the required parabola is \[ \boxed{x^2-8x+8y+24=0} \]
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