The length of the latus rectum of `3x^(2) -4y +6x - 3 = 0` is (a) 3 (b) 2 (c) `4/3` (d) `3/4`

Options

A. 3
B. 2
C. `(4)/(3)`
D. `(3)/(4)`

Explanation

To find the length of the latus rectum of the parabola given by the equation \(3x^2 - 4y + 6x - 3 = 0\), we will first rewrite the equation in standard form.

### Step 1: Rearranging the equation
Start with the given equation:
\[ 3x^2 - 4y + 6x - 3 = 0 \]
Rearranging gives:
\[ 3x^2 + 6x - 4y - 3 = 0 \]

### Step 2: Grouping terms
Next, we will group the \(x\) terms together:
\[ 3(x^2 + 2x) - 4y - 3 = 0 \]

### Step 3: Completing the square
Now, we complete the square for the expression \(x^2 + 2x\):
\[ x^2 + 2x = (x + 1)^2 - 1 \]
Substituting this back into the equation gives:
\[ 3((x + 1)^2 - 1) - 4y - 3 = 0 \]
Simplifying this:
\[ 3(x + 1)^2 - 3 - 4y - 3 = 0 \]
\[ 3(x + 1)^2 - 4y - 6 = 0 \]
\[ 3(x + 1)^2 - 4y = 6 \]

### Step 4: Isolating \(y\)
Now, isolate \(y\):
\[ 4y = 3(x + 1)^2 - 6 \]
\[ y = \frac{3}{4}(x + 1)^2 - \frac{3}{2} \]

### Step 5: Identifying the standard form
This can be written in the standard form of a parabola:
\[ y = \frac{3}{4}(x + 1)^2 - \frac{3}{2} \]
This shows that the parabola opens upwards.

### Step 6: Finding the value of \(a\)
From the standard form \(y = ax^2\), we can identify \(4a = \frac{3}{4}\). Thus:
\[ 4a = \frac{3}{4} \]
To find \(a\):
\[ a = \frac{3}{16} \]

### Step 7: Length of the latus rectum
The length of the latus rectum of a parabola is given by \(4a\):
\[ \text{Length of latus rectum} = 4a = 4 \times \frac{3}{16} = \frac{12}{16} = \frac{3}{4} \]

### Final Answer
The length of the latus rectum is \(\frac{3}{4}\).

### Options
The correct option is (d) \(\frac{3}{4}\).

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