Question:
The vertex of a parabola is at $(2,-5)$ and the focus is at $(5,-5)$. The equation of the parabola is
The vertex of a parabola is at $(2,-5)$ and the focus is at $(5,-5)$. The equation of the parabola is
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If focus is right of vertex, parabola opens right: use $(y-k)^2 = 4a(x-h)$.
Updated On: Apr 24, 2026
- $y^2 + 10y - 10x + 49 = 0$
- $y^2 + 10y - 12x + 49 = 0$
- $y^2 + 10y - 12x + 46 = 0$
- $y^2 + 8y - 12x + 49 = 0$
- $y^2 + 10y - 18x + 48 = 0$
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The Correct Option is B
Solution and Explanation
Concept:
• Standard form: $(y-k)^2 = 4a(x-h)$
• Vertex $(h,k)$, Focus $(h+a,k)$
Step 1: Identify parameters
\[ (h,k) = (2,-5), \quad (h+a,k) = (5,-5) \] \[ a = 3 \]
Step 2: Write equation
\[ (y+5)^2 = 12(x-2) \]
Step 3: Expand
\[ y^2 + 10y + 25 = 12x - 24 \] \[ y^2 + 10y - 12x + 49 = 0 \] Final Conclusion:
Option (B)
• Standard form: $(y-k)^2 = 4a(x-h)$
• Vertex $(h,k)$, Focus $(h+a,k)$
Step 1: Identify parameters
\[ (h,k) = (2,-5), \quad (h+a,k) = (5,-5) \] \[ a = 3 \]
Step 2: Write equation
\[ (y+5)^2 = 12(x-2) \]
Step 3: Expand
\[ y^2 + 10y + 25 = 12x - 24 \] \[ y^2 + 10y - 12x + 49 = 0 \] Final Conclusion:
Option (B)
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