If the tangent and the normal to `x^2-y^2=4` at a point cut off intercepts `a_1,a_2` on the x-axis respectively & `b_1,b_2` on the y-axis respectively. Then the value of `a_1a_2+b_1b_2` is equal to:

Answer

To solve the problem step-by-step, we need to find the intercepts on the x-axis and y-axis for both the tangent and normal to the hyperbola given by the equation \(x^2 - y^2 = 4\).

### Step 1: Identify the hyperbola and its properties
The given equation of the hyperbola is:
\[ x^2 - y^2 = 4 \]
This can be rewritten in standard form as:
\[ \frac{x^2}{4} - \frac{y^2}{4} = 1 \]
From this, we can see that \(a^2 = 4\) and \(b^2 = 4\), which gives us \(a = 2\) and \(b = 2\).

### Step 2: Write the equations for the tangent and normal
For a hyperbola of the form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), the equation of the tangent at a point \((x_1, y_1)\) on the hyperbola is given by:
\[ \frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1 \]
Substituting \(a^2 = 4\) and \(b^2 = 4\), we get the tangent equation:
\[ \frac{xx_1}{4} - \frac{yy_1}{4} = 1 \quad \Rightarrow \quad xx_1 - yy_1 = 4 \]

The equation of the normal at the same point is given by:
\[ \frac{a^2 x}{x_1} + \frac{b^2 y}{y_1} = a^2 - b^2 \]
Since \(a^2 = 4\) and \(b^2 = 4\), we have:
\[ 4\frac{x}{x_1} + 4\frac{y}{y_1} = 0 \quad \Rightarrow \quad x + y = 0 \]

### Step 3: Find the x-intercepts \(a_1\) and \(a_2\)
To find the x-intercept of the tangent, set \(y = 0\):
\[ xx_1 - 0 = 4 \quad \Rightarrow \quad x = \frac{4}{x_1} \]
Thus, the x-intercept \(a_1 = \frac{4}{x_1}\).

To find the x-intercept of the normal, set \(y = 0\):
\[ x + 0 = 0 \quad \Rightarrow \quad x = 0 \]
Thus, the x-intercept \(a_2 = 0\).

### Step 4: Find the y-intercepts \(b_1\) and \(b_2\)
To find the y-intercept of the tangent, set \(x = 0\):
\[ 0 - yy_1 = 4 \quad \Rightarrow \quad y = -\frac{4}{y_1} \]
Thus, the y-intercept \(b_1 = -\frac{4}{y_1}\).

To find the y-intercept of the normal, set \(x = 0\):
\[ 0 + y = 0 \quad \Rightarrow \quad y = 0 \]
Thus, the y-intercept \(b_2 = 0\).

### Step 5: Calculate \(a_1 a_2 + b_1 b_2\)
Now we can calculate:
\[ a_1 a_2 + b_1 b_2 = \left(\frac{4}{x_1}\right)(0) + \left(-\frac{4}{y_1}\right)(0) = 0 + 0 = 0 \]

### Final Answer
Thus, the value of \(a_1 a_2 + b_1 b_2\) is:
\[ \boxed{0} \]