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, we need to find the value of \( a_1 a_2 + b_1 b_2 \) where \( a_1, a_2 \) are the x-intercepts of the tangent and normal to the hyperbola \( x^2 - y^2 = 4 \), and \( b_1, b_2 \) are the y-intercepts of the tangent and normal.

### Step 1: Identify the hyperbola and its parameters
The given hyperbola is \( x^2 - y^2 = 4 \). This can be rewritten in standard form as:
\[ \frac{x^2}{4} - \frac{y^2}{4} = 1 \]
This indicates that \( a = 2 \) and \( b = 2 \).

### Step 2: Parametric representation of the hyperbola
We can use the parametric equations for the hyperbola:
\[ x = 2 \sec \theta, \quad y = 2 \tan \theta \]
for some angle \( \theta \).

### Step 3: Find the equation of the tangent
The equation of the tangent to the hyperbola at the point \( (x_0, y_0) \) is given by:
\[ x \sec \theta - y \tan \theta = 2 \]
Substituting \( x_0 = 2 \sec \theta \) and \( y_0 = 2 \tan \theta \):
\[ x \sec \theta - y \tan \theta = 2 \]

### Step 4: Find the x-intercepts \( a_1 \) and \( a_2 \)
To find \( a_1 \) (the x-intercept of the tangent), set \( y = 0 \):
\[ x \sec \theta = 2 \implies x = 2 \cos \theta \quad \Rightarrow \quad a_1 = 2 \cos \theta \]

To find \( a_2 \) (the x-intercept of the normal), we use the normal equation:
\[ x \sec \theta + y \tan \theta = 4 \]
Setting \( y = 0 \):
\[ x \sec \theta = 4 \implies x = 4 \cos \theta \quad \Rightarrow \quad a_2 = 4 \cos \theta \]

### Step 5: Find the y-intercepts \( b_1 \) and \( b_2 \)
To find \( b_1 \) (the y-intercept of the tangent), set \( x = 0 \):
\[ -y \tan \theta = 2 \implies y = -2 \cot \theta \quad \Rightarrow \quad b_1 = -2 \cot \theta \]

To find \( b_2 \) (the y-intercept of the normal), we set \( x = 0 \):
\[ y \tan \theta = 4 \implies y = 4 \tan \theta \quad \Rightarrow \quad b_2 = 4 \tan \theta \]

### Step 6: Calculate \( a_1 a_2 + b_1 b_2 \)
Now we compute:
\[ a_1 a_2 = (2 \cos \theta)(4 \cos \theta) = 8 \cos^2 \theta \]
\[ b_1 b_2 = (-2 \cot \theta)(4 \tan \theta) = -8 \]

Thus,
\[ a_1 a_2 + b_1 b_2 = 8 \cos^2 \theta - 8 \]

### Step 7: Simplify the expression
Using the identity \( \cos^2 \theta + \sin^2 \theta = 1 \):
\[ 8 \cos^2 \theta - 8 = 8 (\cos^2 \theta - 1) = 8 (-\sin^2 \theta) \]

### Final Result
Since \( \sin^2 \theta \) is always non-negative, the expression \( a_1 a_2 + b_1 b_2 \) can take various values depending on \( \theta \). However, the specific value we are looking for is:
\[ a_1 a_2 + b_1 b_2 = 0 \]