Hyperbola `x^2/a^2-y^2/b^2=2` has eccentricity `5/4`. If normal to the hyperbola at `(8/sqrt5,12/5)` is `8sqrt5x+betay=lamda` then the value of `(lamda-beta)` is

Answer

To solve the problem step by step, we will follow the given information about the hyperbola and the normal at the specified point.

### Step 1: Understand the Hyperbola
The equation of the hyperbola is given as:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 2 \]
From this, we can rewrite it as:
\[ x^2 - \frac{a^2}{b^2}y^2 = 2a^2 \]

### Step 2: Find the Eccentricity
The eccentricity \( e \) of the hyperbola is given by:
\[ e = \sqrt{1 + \frac{b^2}{a^2}} \]
We know that \( e = \frac{5}{4} \). Squaring both sides gives:
\[ \left(\frac{5}{4}\right)^2 = 1 + \frac{b^2}{a^2} \]
\[ \frac{25}{16} = 1 + \frac{b^2}{a^2} \]
Subtracting 1 from both sides:
\[ \frac{25}{16} - 1 = \frac{b^2}{a^2} \]
\[ \frac{25}{16} - \frac{16}{16} = \frac{b^2}{a^2} \]
\[ \frac{9}{16} = \frac{b^2}{a^2} \]
Thus, we have:
\[ b^2 = \frac{9}{16} a^2 \]

### Step 3: Find the Normal Equation
The normal to the hyperbola at a point \( (x_1, y_1) \) is given by:
\[ \frac{a^2 x}{x_1} + \frac{b^2 y}{y_1} = a^2 e^2 \]
Given point is \( \left(\frac{8}{\sqrt{5}}, \frac{12}{5}\right) \).

Substituting \( x_1 = \frac{8}{\sqrt{5}} \) and \( y_1 = \frac{12}{5} \):
\[ \frac{a^2 x}{\frac{8}{\sqrt{5}}} + \frac{b^2 y}{\frac{12}{5}} = a^2 \left(\frac{5}{4}\right)^2 \]
\[ \frac{a^2 \sqrt{5} x}{8} + \frac{b^2 \cdot 5y}{12} = \frac{25}{16} a^2 \]

### Step 4: Substitute \( b^2 \)
Substituting \( b^2 = \frac{9}{16} a^2 \):
\[ \frac{a^2 \sqrt{5} x}{8} + \frac{\frac{9}{16} a^2 \cdot 5y}{12} = \frac{25}{16} a^2 \]
Multiplying through by \( 16 \) to eliminate the denominators:
\[ 2 a^2 \sqrt{5} x + \frac{15}{12} a^2 y = 25 a^2 \]
This simplifies to:
\[ 2 \sqrt{5} x + \frac{15}{12} y = 25 \]

### Step 5: Rearranging the Normal Equation
Rearranging gives:
\[ 2 \sqrt{5} x + \frac{15}{12} y = 25 \]
This can be expressed in the form:
\[ 8\sqrt{5} x + \beta y = \lambda \]
where \( \beta = \frac{15}{12} \) and \( \lambda = 25 \).

### Step 6: Calculate \( \lambda - \beta \)
Now we need to find \( \lambda - \beta \):
\[ \lambda = 25, \quad \beta = \frac{15}{12} = \frac{5}{4} \]
Thus,
\[ \lambda - \beta = 25 - \frac{5}{4} \]
Converting 25 to a fraction:
\[ 25 = \frac{100}{4} \]
So,
\[ \lambda - \beta = \frac{100}{4} - \frac{5}{4} = \frac{95}{4} \]

### Final Answer
The value of \( \lambda - \beta \) is:
\[ \frac{95}{4} = 23.75 \]