Consider the parabola whose focus is at (0,0) and tangent at vertex is` x-y+1=0`
The length of latus rectum is (a) `4sqrt(2)` (b) `2sqrt(2)` (c) `8sqrt(2)` (d) `3sqrt(2)`

Options

A. `4sqrt(2)`
B. `2sqrt(2)`
C. `8sqrt(2)`
D. `3sqrt(2)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the standard form of the parabola
The standard equation of a parabola with focus at (0, 0) is given by:
\[ y^2 = 4ax \]
where \(a\) is the distance from the vertex to the focus.

### Step 2: Analyze the given tangent line
The tangent line at the vertex is given by:
\[ x - y + 1 = 0 \]
We can rewrite this in the standard form \(Ax + By + C = 0\):
\[ 1x - 1y + 1 = 0 \]
Here, \(A = 1\), \(B = -1\), and \(C = 1\).

### Step 3: Calculate the distance from the focus to the tangent line
The distance \(d\) from a point \((x_1, y_1)\) to a line \(Ax + By + C = 0\) is given by:
\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
Substituting the coordinates of the focus \((0, 0)\) into the formula:
\[ d = \frac{|1(0) - 1(0) + 1|}{\sqrt{1^2 + (-1)^2}} = \frac{|1|}{\sqrt{1 + 1}} = \frac{1}{\sqrt{2}} \]

### Step 4: Relate the distance to the parameter \(a\)
Since the distance from the vertex to the focus is \(a\), we have:
\[ a = \frac{1}{\sqrt{2}} \]

### Step 5: Calculate the length of the latus rectum
The length of the latus rectum \(L\) of a parabola is given by:
\[ L = 4a \]
Substituting the value of \(a\):
\[ L = 4 \times \frac{1}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2} \]

### Step 6: Conclusion
The length of the latus rectum is:
\[ \boxed{2\sqrt{2}} \]