If the locus of the image of the point `(lambda^(2), 2lambda)` in the line mirror `x-y+1=0` (where `lambda` is a parameter) is
`(x-a)^(2)=b(y-c)` where `a, b , c in I`, then the value of `((a+b)/(c+b))` is equal to

Answer

To solve the problem, we need to find the locus of the image of the point \((\lambda^2, 2\lambda)\) when reflected across the line \(x - y + 1 = 0\).

### Step 1: Identify the line and its slope
The line \(x - y + 1 = 0\) can be rewritten in slope-intercept form as \(y = x + 1\). The slope of this line, \(m_1\), is 1.

**Hint:** The slope of a line in the form \(Ax + By + C = 0\) can be found by rewriting it in the form \(y = mx + c\).

### Step 2: Set up the point and its image
Let \(P = (\lambda^2, 2\lambda)\) be the point. Let \(Q = (h, k)\) be the image of point \(P\) after reflection across the line.

**Hint:** The image point \(Q\) will satisfy certain conditions based on the properties of reflection.

### Step 3: Use the property of perpendicular slopes
The slope of the line segment \(PQ\) (from \(P\) to \(Q\)) is given by:
\[ m_2 = \frac{k - 2\lambda}{h - \lambda^2} \]
Since the line \(x - y + 1 = 0\) and the line segment \(PQ\) are perpendicular, we have:
\[ m_1 \cdot m_2 = -1 \implies 1 \cdot \frac{k - 2\lambda}{h - \lambda^2} = -1 \]
This simplifies to:
\[ k - 2\lambda = - (h - \lambda^2) \implies k - 2\lambda = -h + \lambda^2 \]
Rearranging gives:
\[ h + k = \lambda^2 + 2\lambda \]

**Hint:** Remember that the product of the slopes of two perpendicular lines is \(-1\).

### Step 4: Find the midpoint of segment \(PQ\)
The midpoint \(R\) of segment \(PQ\) is given by:
\[ R = \left(\frac{\lambda^2 + h}{2}, \frac{2\lambda + k}{2}\right) \]
This midpoint must lie on the line \(x - y + 1 = 0\), leading to:
\[ \frac{\lambda^2 + h}{2} - \frac{2\lambda + k}{2} + 1 = 0 \]
Multiplying through by 2 gives:
\[ \lambda^2 + h - 2\lambda - k + 2 = 0 \]
Rearranging this yields:
\[ h - k = -\lambda^2 + 2\lambda - 2 \]

**Hint:** The midpoint of a segment can be used to find relationships between the coordinates of the endpoints.

### Step 5: Solve for \(h\) and \(k\)
From the equations:
1. \(h + k = \lambda^2 + 2\lambda\)
2. \(h - k = -\lambda^2 + 2\lambda - 2\)

Adding these two equations:
\[ 2h = (\lambda^2 + 2\lambda) + (-\lambda^2 + 2\lambda - 2) = 4\lambda - 2 \]
Thus,
\[ h = 2\lambda - 1 \]

Now substituting \(h\) back to find \(k\):
\[ k = \lambda^2 + 2\lambda - h = \lambda^2 + 2\lambda - (2\lambda - 1) = \lambda^2 + 1 \]

**Hint:** Use simultaneous equations to find the values of unknowns.

### Step 6: Establish the locus equation
We have:
\[ h = 2\lambda - 1 \quad \text{and} \quad k = \lambda^2 + 1 \]
Substituting \(h\) and \(k\) into the equation:
\[ (2\lambda - 1) + 1 = 2\lambda \quad \text{and} \quad k - 1 = \lambda^2 \]
This gives:
\[ h + 1 = 2\lambda \quad \text{and} \quad k - 1 = \lambda^2 \]
Thus, we can write:
\[ (h + 1)^2 = 4(k - 1) \]

### Step 7: Compare with the standard form
Rearranging gives:
\[ (h + 1)^2 = 4(k - 1) \]
This matches the form \((x - a)^2 = b(y - c)\) where:
- \(a = -1\)
- \(b = 4\)
- \(c = 1\)

### Step 8: Calculate \(\frac{a + b}{c + b}\)
Now we calculate:
\[ \frac{a + b}{c + b} = \frac{-1 + 4}{1 + 4} = \frac{3}{5} \]

Thus, the final answer is:
\[ \frac{3}{5} \]