The length of the latus rectum of the parabola `x^2 - 6x + 5y = 0` is

Options

A. 3
B. 5
C. 7
D. 1

Answer

To find the length of the latus rectum of the parabola given by the equation \(x^2 - 6x + 5y = 0\), we can follow these steps:

### Step 1: Rearrange the equation
Start by rearranging the given equation to isolate \(y\):
\[ x^2 - 6x + 5y = 0 \implies 5y = -x^2 + 6x \implies y = -\frac{1}{5}x^2 + \frac{6}{5}x \]

### Step 2: Complete the square
Next, we will complete the square for the \(x\) terms:
\[ y = -\frac{1}{5}(x^2 - 6x) \]
To complete the square, take half of the coefficient of \(x\) (which is \(-6\)), square it, and add/subtract it inside the parentheses:
\[ y = -\frac{1}{5}((x - 3)^2 - 9) = -\frac{1}{5}(x - 3)^2 + \frac{9}{5} \]

### Step 3: Rewrite the equation in standard form
Now, we can rewrite the equation in the standard form of a parabola:
\[ y - \frac{9}{5} = -\frac{1}{5}(x - 3)^2 \]
This can be rearranged to:
\[ (x - 3)^2 = -5(y - \frac{9}{5}) \]
This is in the form \((x - h)^2 = -4a(y - k)\), where \((h, k)\) is the vertex of the parabola.

### Step 4: Identify parameters
From the equation \((x - 3)^2 = -4a(y - \frac{9}{5})\), we can identify:
- \(h = 3\)
- \(k = \frac{9}{5}\)
- \(4a = 5\) (since \(-4a = -5\))

### Step 5: Solve for \(a\)
Now, we can solve for \(a\):
\[ 4a = 5 \implies a = \frac{5}{4} \]

### Step 6: Calculate the length of the latus rectum
The length of the latus rectum \(L\) of a parabola is given by the formula:
\[ L = 4a \]
Substituting the value of \(a\):
\[ L = 4 \times \frac{5}{4} = 5 \]

### Final Answer
Thus, the length of the latus rectum of the parabola is \(5\).

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