`L_(1)` is a line passing through the point `A(n,n+1)` having slope, `n, L_(2)` is a line passing through the point `B(-n, n^(2))` and is perpendicular to `L_(1)`, If `L_(1)` and `L_(2)` intersect on y-axis, then n is equal to ________ `(n gt 0)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the equations of the lines
We know that line \( L_1 \) passes through point \( A(n, n+1) \) and has a slope of \( n \). The equation of a line in point-slope form is given by:

\[ y - y_1 = m(x - x_1) \]

For line \( L_1 \):

\[ y - (n + 1) = n(x - n) \]

Expanding this, we get:

\[ y - n - 1 = nx - n^2 \]
\[ y = nx - n^2 + n + 1 \]

### Step 2: Find the slope of line \( L_2 \)
Since line \( L_2 \) is perpendicular to line \( L_1 \), its slope \( m_2 \) is the negative reciprocal of \( n \):

\[ m_2 = -\frac{1}{n} \]

Line \( L_2 \) passes through point \( B(-n, n^2) \). Using the point-slope form for line \( L_2 \):

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

Expanding this, we get:

\[ y - n^2 = -\frac{1}{n}x - 1 \]
\[ y = -\frac{1}{n}x + n^2 - 1 \]

### Step 3: Find the intersection on the y-axis
To find where these two lines intersect on the y-axis, we set \( x = 0 \) in both equations.

For \( L_1 \):

\[ y = n(0) - n^2 + n + 1 = -n^2 + n + 1 \]

For \( L_2 \):

\[ y = -\frac{1}{n}(0) + n^2 - 1 = n^2 - 1 \]

### Step 4: Set the y-values equal to find \( n \)
Since the lines intersect at the same point on the y-axis, we set the two equations for \( y \) equal to each other:

\[ -n^2 + n + 1 = n^2 - 1 \]

### Step 5: Rearranging the equation
Rearranging gives:

\[ -n^2 - n^2 + n + 1 + 1 = 0 \]
\[ -2n^2 + n + 2 = 0 \]

Multiplying through by -1:

\[ 2n^2 - n - 2 = 0 \]

### Step 6: Solve the quadratic equation
Using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = 2, b = -1, c = -2 \):

\[ n = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 2 \cdot (-2)}}{2 \cdot 2} \]
\[ n = \frac{1 \pm \sqrt{1 + 16}}{4} \]
\[ n = \frac{1 \pm \sqrt{17}}{4} \]

### Step 7: Determine the valid solution
Since \( n > 0 \), we take the positive root:

\[ n = \frac{1 + \sqrt{17}}{4} \]

### Final Answer
Thus, the value of \( n \) is:

\[ n = \frac{1 + \sqrt{17}}{4} \]

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