Question:medium

The equation of the parabola with focus \((2,0)\) and vertex \((1,0)\) is

Show Hint

For a parabola opening right with vertex \((h,k)\), use \((y-k)^2=4a(x-h)\).
  • \(y^2=4x\)
  • \(y^2=4x-4\)
  • \(y^2=4(x+1)\)
  • \(y^2=-4(x-1)\)
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
We need to find the equation of a parabola given its vertex and focus. The position of the focus relative to the vertex determines the orientation of the parabola and the value of the parameter 'a'.
Step 2: Key Formula or Approach:
The standard equation of a parabola that opens horizontally with vertex at $(h, k)$ is: \[ (y-k)^2 = 4a(x-h) \] The focus is at $(h+a, k)$ and the directrix is the line $x = h-a$. The parameter 'a' is the distance from the vertex to the focus.
Step 3: Detailed Explanation:
We are given:
Vertex $(h, k) = (1, 0)$
Focus $= (2, 0)$
From the coordinates, we can see that the vertex and focus lie on the x-axis ($y=0$), so the axis of the parabola is the x-axis. Since the focus $(2,0)$ is to the right of the vertex $(1,0)$, the parabola opens to the right. This means 'a' will be positive. The distance 'a' from the vertex to the focus is the difference in their x-coordinates: \[ a = (\text{x-coordinate of focus}) - (\text{x-coordinate of vertex}) = 2 - 1 = 1 \] Now, substitute the values of the vertex $(h, k) = (1, 0)$ and $a=1$ into the standard equation: \[ (y-k)^2 = 4a(x-h) \] \[ (y-0)^2 = 4(1)(x-1) \] \[ y^2 = 4(x-1) \] Expanding this equation gives: \[ y^2 = 4x - 4 \] Step 4: Final Answer:
The equation of the parabola is $y^2 = 4x - 4$. Therefore, option (B) is correct.
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