Show that the parametric point `(2+t^(2),2t+1)` represents a parabola. Show that its vertex is (2,1).

Options

A. a parabola with focus at (2, 1)
B. a parabola with vertex at (2, 1)
C. an ellipse with centre at (2, 1)
D. none of these

Answer

To show that the parametric point \((2+t^2, 2t+1)\) represents a parabola and to find its vertex, we can follow these steps:

### Step 1: Define the Parametric Equations
Let:
\[ x = 2 + t^2 \]
\[ y = 2t + 1 \]

### Step 2: Solve for \(t\) in terms of \(y\)
From the equation for \(y\):
\[ y = 2t + 1 \implies 2t = y - 1 \implies t = \frac{y - 1}{2} \]

### Step 3: Substitute \(t\) into the equation for \(x\)
Now, substitute \(t = \frac{y - 1}{2}\) into the equation for \(x\):
\[ x = 2 + t^2 = 2 + \left(\frac{y - 1}{2}\right)^2 \]
Calculating \(\left(\frac{y - 1}{2}\right)^2\):
\[ \left(\frac{y - 1}{2}\right)^2 = \frac{(y - 1)^2}{4} \]
Thus,
\[ x = 2 + \frac{(y - 1)^2}{4} \]

### Step 4: Rearranging the Equation
Rearranging the equation gives:
\[ x - 2 = \frac{(y - 1)^2}{4} \]
Multiplying both sides by 4:
\[ 4(x - 2) = (y - 1)^2 \]

### Step 5: Identify the Parabola
This equation can be rewritten as:
\[ (y - 1)^2 = 4(x - 2) \]
This is the standard form of a parabola that opens to the right with vertex at \((2, 1)\).

### Step 6: Conclusion
Thus, we have shown that the parametric point \((2+t^2, 2t+1)\) represents a parabola, and its vertex is at \((2, 1)\).

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