An elipse has it centre at `(1,-1)` and semimajor axis `=8` and which passes through the point `(1,3)` if l be length of its latus rectum then finr `l/4.`

Answer

To solve the problem step by step, we will follow the information given about the ellipse and derive the required values.

### Step 1: Write the equation of the ellipse
The standard form of the equation of an ellipse centered at \((h, k)\) is given by:

\[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \]

Given:
- Center \((h, k) = (1, -1)\)
- Semi-major axis \(a = 8\)

Substituting these values into the equation, we have:

\[ \frac{(x-1)^2}{8^2} + \frac{(y+1)^2}{b^2} = 1 \]

This simplifies to:

\[ \frac{(x-1)^2}{64} + \frac{(y+1)^2}{b^2} = 1 \]

### Step 2: Use the point (1, 3) to find b
Since the ellipse passes through the point \((1, 3)\), we can substitute \(x = 1\) and \(y = 3\) into the equation to find \(b\).

Substituting:

\[ \frac{(1-1)^2}{64} + \frac{(3+1)^2}{b^2} = 1 \]

This simplifies to:

\[ 0 + \frac{4^2}{b^2} = 1 \]

Thus:

\[ \frac{16}{b^2} = 1 \]

### Step 3: Solve for b
From the equation \(\frac{16}{b^2} = 1\), we can cross-multiply to find \(b^2\):

\[ 16 = b^2 \]

Taking the square root gives:

\[ b = 4 \]

### Step 4: Calculate the length of the latus rectum
The length of the latus rectum \(L\) of an ellipse is given by the formula:

\[ L = \frac{2b^2}{a} \]

Substituting the values of \(b\) and \(a\):

\[ L = \frac{2 \times 4^2}{8} = \frac{2 \times 16}{8} = \frac{32}{8} = 4 \]

### Step 5: Find \(L/4\)
Now, we need to find \(\frac{L}{4}\):

\[ \frac{L}{4} = \frac{4}{4} = 1 \]

### Final Answer
Thus, the value of \(\frac{L}{4}\) is:

\[ \boxed{1} \]