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 derive the equation of the ellipse and find the required value of \( \frac{L}{4} \).

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

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

Here, we are given the center of the ellipse as \((1, -1)\), so \(h = 1\) and \(k = -1\).

### Step 2: Substitute the center into the equation
Substituting \(h\) and \(k\) into the equation, we get:

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

### Step 3: Identify the semi-major axis
We are given that the semi-major axis \(a = 8\). Therefore, \(a^2 = 8^2 = 64\).

### Step 4: Substitute the value of \(a^2\) into the equation
Now, substituting \(a^2\) into the equation, we have:

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

### Step 5: Use the point (1, 3) to find \(b^2\)
The ellipse passes through the point \((1, 3)\). We will substitute \(x = 1\) and \(y = 3\) into the equation:

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

This simplifies to:

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

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

### Step 6: Solve for \(b^2\)
From the above equation, we can solve for \(b^2\):

\[ b^2 = 16 \]

### Step 7: Find the length of the latus rectum \(L\)
The length of the latus rectum \(L\) of an ellipse is given by the formula:

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

Substituting the values of \(b^2\) and \(a\):

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

### Step 8: Find \(\frac{L}{4}\)
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:

\[ \boxed{1} \]