If the tangent and normal to the rectangular hyperbola xy=16 at point (8,2) cut off intercepts `a_(1),a_(2)` on the x axis and `b_(1)b_(2)` on the y axis then `a_(1)a_(2)+b_(1)b_(2)+7//2` equals

Answer

To solve the problem, we need to find the intercepts on the x-axis and y-axis from the tangent and normal lines to the hyperbola \( xy = 16 \) at the point \( (8, 2) \).

### Step 1: Find the slope of the tangent line
The equation of the hyperbola is given by \( xy = 16 \). To find the slope of the tangent at the point \( (8, 2) \), we first differentiate the equation implicitly.

1. Differentiate \( xy = 16 \):
\[ x \frac{dy}{dx} + y = 0 \implies \frac{dy}{dx} = -\frac{y}{x} \]

2. Substitute \( x = 8 \) and \( y = 2 \):
\[ \frac{dy}{dx} = -\frac{2}{8} = -\frac{1}{4} \]

### Step 2: Write the equation of the tangent line
Using the point-slope form of the line:
\[ y - y_1 = m(x - x_1) \]
Substituting \( m = -\frac{1}{4} \), \( x_1 = 8 \), and \( y_1 = 2 \):
\[ y - 2 = -\frac{1}{4}(x - 8) \]
Rearranging gives:
\[ y - 2 = -\frac{1}{4}x + 2 \implies \frac{1}{4}x + y = 4 \implies x + 4y = 16 \]

### Step 3: Find the x-intercepts and y-intercepts of the tangent line
1. **X-intercept** (\( y = 0 \)):
\[ x + 4(0) = 16 \implies x = 16 \quad \text{(let's denote this as } a_1) \]

2. **Y-intercept** (\( x = 0 \)):
\[ 0 + 4y = 16 \implies y = 4 \quad \text{(let's denote this as } b_1) \]

### Step 4: Find the equation of the normal line
The slope of the normal line is the negative reciprocal of the slope of the tangent:
\[ \text{slope of normal} = 4 \]
Using the point-slope form again:
\[ y - 2 = 4(x - 8) \]
Rearranging gives:
\[ y - 2 = 4x - 32 \implies y = 4x - 30 \]

### Step 5: Find the x-intercepts and y-intercepts of the normal line
1. **X-intercept** (\( y = 0 \)):
\[ 0 = 4x - 30 \implies 4x = 30 \implies x = \frac{15}{2} \quad \text{(let's denote this as } a_2) \]

2. **Y-intercept** (\( x = 0 \)):
\[ y = 4(0) - 30 = -30 \quad \text{(let's denote this as } b_2) \]

### Step 6: Calculate \( a_1 a_2 + b_1 b_2 + \frac{7}{2} \)
Now we have:
- \( a_1 = 16 \)
- \( a_2 = \frac{15}{2} \)
- \( b_1 = 4 \)
- \( b_2 = -30 \)

Calculating \( a_1 a_2 \):
\[ a_1 a_2 = 16 \cdot \frac{15}{2} = 120 \]

Calculating \( b_1 b_2 \):
\[ b_1 b_2 = 4 \cdot (-30) = -120 \]

Now substituting into the expression:
\[ a_1 a_2 + b_1 b_2 + \frac{7}{2} = 120 - 120 + \frac{7}{2} = \frac{7}{2} \]

### Final Answer
Thus, the value of \( a_1 a_2 + b_1 b_2 + \frac{7}{2} \) is:
\[ \frac{7}{2} \]