Equation of the parabola whose axis is parallel to Y- axis and which passes through the point (1,0),(0,0)and (-2,4) , is

Options

A. `2x^2+2y=3y`
B. `2x^2-2x=3y`
C. `2x^2+2x=y`
D. `2x^2-2x=y`

Answer

To find the equation of the parabola whose axis is parallel to the Y-axis and which passes through the points (1, 0), (0, 0), and (-2, 4), we can follow these steps:

### Step 1: Write the general form of the parabola
Since the axis of the parabola is parallel to the Y-axis, we can express the equation of the parabola in the form:
\[ y = ax^2 + bx + c \]

### Step 2: Substitute the points into the equation
We will substitute the given points into the equation to form a system of equations.

1. For the point (0, 0):
\[ 0 = a(0)^2 + b(0) + c \]
This simplifies to:
\[ c = 0 \]

2. For the point (1, 0):
\[ 0 = a(1)^2 + b(1) + c \]
Substituting \( c = 0 \):
\[ 0 = a + b \]
This gives us our first equation:
\[ a + b = 0 \quad \text{(1)} \]

3. For the point (-2, 4):
\[ 4 = a(-2)^2 + b(-2) + c \]
Substituting \( c = 0 \):
\[ 4 = 4a - 2b \]
This gives us our second equation:
\[ 4 = 4a - 2b \quad \text{(2)} \]

### Step 3: Solve the system of equations
From equation (1), we can express \( b \) in terms of \( a \):
\[ b = -a \]

Now, substitute \( b = -a \) into equation (2):
\[ 4 = 4a - 2(-a) \]
\[ 4 = 4a + 2a \]
\[ 4 = 6a \]
\[ a = \frac{4}{6} = \frac{2}{3} \]

Now, substitute \( a \) back into equation (1) to find \( b \):
\[ a + b = 0 \]
\[ \frac{2}{3} + b = 0 \]
\[ b = -\frac{2}{3} \]

### Step 4: Write the final equation of the parabola
Now that we have \( a \), \( b \), and \( c \):
- \( a = \frac{2}{3} \)
- \( b = -\frac{2}{3} \)
- \( c = 0 \)

The equation of the parabola is:
\[ y = \frac{2}{3}x^2 - \frac{2}{3}x \]

### Step 5: Rearranging the equation
To express it in a standard form, we can multiply through by 3 to eliminate the fractions:
\[ 3y = 2x^2 - 2x \]

Thus, the equation of the parabola is:
\[ 2x^2 - 2x - 3y = 0 \]