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 find the directrix, equation of the axis, and length of the latus rectum, we can follow these steps:

### Step 1: Rearranging the equation
We start with the equation:
\[ 2y^2 + 3y - 4x - 3 = 0 \]
Rearranging it gives:
\[ 2y^2 + 3y = 4x + 3 \]

### Step 2: Completing the square for \(y\)
To complete the square for the \(y\) terms, we can factor out the coefficient of \(y^2\):
\[ 2\left(y^2 + \frac{3}{2}y\right) = 4x + 3 \]
Now, we complete the square inside the parentheses:
\[ y^2 + \frac{3}{2}y = \left(y + \frac{3}{4}\right)^2 - \frac{9}{16} \]
So, substituting back, we have:
\[ 2\left(\left(y + \frac{3}{4}\right)^2 - \frac{9}{16}\right) = 4x + 3 \]
This simplifies to:
\[ 2\left(y + \frac{3}{4}\right)^2 - \frac{9}{8} = 4x + 3 \]
Adding \(\frac{9}{8}\) to both sides gives:
\[ 2\left(y + \frac{3}{4}\right)^2 = 4x + 3 + \frac{9}{8} \]
Converting \(3\) to eighths:
\[ 3 = \frac{24}{8} \]
So, we have:
\[ 2\left(y + \frac{3}{4}\right)^2 = 4x + \frac{33}{8} \]

### Step 3: Dividing by 2
Now, divide the entire equation by 2:
\[ \left(y + \frac{3}{4}\right)^2 = 2x + \frac{33}{16} \]
Rearranging gives:
\[ \left(y + \frac{3}{4}\right)^2 = 2\left(x + \frac{33}{32}\right) \]

### Step 4: Identifying the vertex
From the equation \(\left(y - k\right)^2 = 4p\left(x - h\right)\), we can identify:
- Vertex \((h, k) = \left(-\frac{33}{32}, -\frac{3}{4}\right)\)

### Step 5: Finding the axis of symmetry
The axis of symmetry is given by the line:
\[ y + \frac{3}{4} = 0 \quad \Rightarrow \quad y = -\frac{3}{4} \]

### Step 6: Finding the directrix
The value of \(p\) (which is \(a\) in the standard form) is:
\[ p = \frac{1}{2} \]
The directrix is given by \(x = h - p\):
\[ x = -\frac{33}{32} - \frac{1}{2} = -\frac{33}{32} - \frac{16}{32} = -\frac{49}{32} \]

### Step 7: Finding the length of the latus rectum
The length of the latus rectum is given by:
\[ 4p = 4 \times \frac{1}{2} = 2 \]

### Final Results
- **Vertex**: \(\left(-\frac{33}{32}, -\frac{3}{4}\right)\)
- **Equation of the Axis**: \(y = -\frac{3}{4}\)
- **Directrix**: \(x = -\frac{49}{32}\)
- **Length of Latus Rectum**: \(2\)