Let V be the vertex and L be the latusrectum of the parabola `x^2=2y+4x-4`. Then the equation of the parabola whose vertex is at V. Latusrectum `L//2` and axis s perpendicular to the axis of the given parabola.

Options

A. `y^2=x-2`
B. `y^2=x-4`
C. `y^2=2-x`
D. `y^2=4-x`

Answer

To solve the problem step by step, we will follow these steps:

### Step 1: Rewrite the given equation of the parabola
The given equation is:
\[ x^2 = 2y + 4x - 4 \]

### Step 2: Rearrange the equation
Rearranging the equation, we have:
\[ x^2 - 4x + 4 = 2y \]

### Step 3: Complete the square
Completing the square for the left-hand side:
\[ (x - 2)^2 = 2y \]

### Step 4: Identify the vertex
From the equation \((x - 2)^2 = 2y\), we can see that the vertex \(V\) of the parabola is at:
\[ V(2, 0) \]

### Step 5: Find the latus rectum of the given parabola
The standard form of the parabola \((x - h)^2 = 4p(y - k)\) gives us the length of the latus rectum \(L\) as \(4p\). Here, \(4p = 2\), so:
\[ p = \frac{1}{2} \]
Thus, the length of the latus rectum \(L\) is:
\[ L = 2p = 2 \]

### Step 6: Determine the new latus rectum
The new latus rectum \(L_2\) is half of the original latus rectum:
\[ L_2 = \frac{L}{2} = \frac{2}{2} = 1 \]

### Step 7: Set up the equation of the new parabola
Since the new parabola has its vertex at \(V(2, 0)\) and its axis is perpendicular to the original parabola, we will use the form:
\[ (y - k)^2 = 4p(x - h) \]
Substituting \(h = 2\), \(k = 0\), and \(4p = 1\) (since \(L_2 = 1\)):
\[ (y - 0)^2 = 1(x - 2) \]
or
\[ y^2 = x - 2 \]

### Step 8: Identify the correct form
The equation can be rearranged to:
\[ y^2 = x - 2 \]

### Step 9: Check the options
Now we check the options provided:
1. \(y^2 = x - 2\)
2. \(y^2 = x - 4\)
3. \(y^2 = 2 - x\)
4. \(y^2 = 4 - x\)

The correct answer is:
\[ y^2 = x - 2 \]

### Final Answer
The equation of the parabola whose vertex is at \(V\), latus rectum \(L/2\), and axis perpendicular to the axis of the given parabola is:
\[ y^2 = x - 2 \]

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