The equation `2y^(2)+3y-4x-3=0` represents a parabola for which

Options

A. length of latus rectum is 4
B. equation of the axis is 4y + 3 = 0
C. equation of directrix is 2x + 1 = 0
D. equation of tangent at vertex is `x=-33/32`

Answer

To solve the equation \(2y^2 + 3y - 4x - 3 = 0\) and analyze the properties of the parabola it represents, we will follow these steps:

### Step 1: Rearranging the Equation
We start with the given equation:
\[ 2y^2 + 3y - 4x - 3 = 0 \]
We can rearrange this to isolate \(x\):
\[ 4x = 2y^2 + 3y - 3 \]
\[ x = \frac{1}{4}(2y^2 + 3y - 3) \]

### Step 2: Factor Out the Coefficient of \(y^2\)
Next, we factor out the coefficient of \(y^2\) (which is 2) from the equation:
\[ x = \frac{1}{4}(2(y^2 + \frac{3}{2}y) - 3) \]
\[ x = \frac{1}{2}(y^2 + \frac{3}{2}y) - \frac{3}{4} \]

### Step 3: Completing the Square
Now, we complete the square for the expression \(y^2 + \frac{3}{2}y\):
\[ y^2 + \frac{3}{2}y = \left(y + \frac{3}{4}\right)^2 - \left(\frac{3}{4}\right)^2 \]
\[ = \left(y + \frac{3}{4}\right)^2 - \frac{9}{16} \]
Substituting this back into the equation for \(x\):
\[ x = \frac{1}{2}\left(\left(y + \frac{3}{4}\right)^2 - \frac{9}{16}\right) - \frac{3}{4} \]
\[ = \frac{1}{2}\left(y + \frac{3}{4}\right)^2 - \frac{9}{32} - \frac{24}{32} \]
\[ = \frac{1}{2}\left(y + \frac{3}{4}\right)^2 - \frac{33}{32} \]

### Step 4: Identifying the Vertex
Now, we can express the parabola in standard form:
\[ x + \frac{33}{32} = \frac{1}{2}\left(y + \frac{3}{4}\right)^2 \]
From this, we can identify the vertex of the parabola:
\[ \text{Vertex} = \left(-\frac{33}{32}, -\frac{3}{4}\right) \]

### Step 5: Finding the Length of the Latus Rectum
The standard form of a parabola is given by \(x = a(y - k)^2 + h\). Here, \(a = \frac{1}{2}\). The length of the latus rectum is given by:
\[ \text{Length of Latus Rectum} = \frac{4}{|a|} = \frac{4}{\frac{1}{2}} = 8 \]

### Step 6: Finding the Equation of the Axis
The axis of the parabola is a horizontal line through the vertex:
\[ y = -\frac{3}{4} \]

### Step 7: Finding the Equation of the Directrix
The directrix is given by:
\[ x = h - \frac{1}{4a} \]
Substituting \(h = -\frac{33}{32}\) and \(a = \frac{1}{2}\):
\[ x = -\frac{33}{32} - \frac{1}{2} = -\frac{33}{32} - \frac{16}{32} = -\frac{49}{32} \]

### Step 8: Finding the Equation of the Tangent at the Vertex
The equation of the tangent at the vertex (which is vertical) is:
\[ x = -\frac{33}{32} \]

### Summary of Results
1. **Length of Latus Rectum**: 8
2. **Equation of Axis**: \(y = -\frac{3}{4}\)
3. **Equation of Directrix**: \(x = -\frac{49}{32}\)
4. **Equation of Tangent at Vertex**: \(x = -\frac{33}{32}\)