To solve the problem step by step, we will follow the outlined steps below:
### Step 1: Identify the point P on the hyperbola
The equation of the hyperbola is given by:
\[
\frac{x^2}{16} - \frac{y^2}{9} = 1
\]
We know the abscissa (x-coordinate) of point P is 5. We can substitute \( x = 5 \) into the hyperbola equation to find the corresponding y-coordinate.
\[
\frac{5^2}{16} - \frac{y^2}{9} = 1
\]
Calculating \( \frac{25}{16} \):
\[
\frac{25}{16} - \frac{y^2}{9} = 1
\]
Rearranging gives:
\[
\frac{y^2}{9} = \frac{25}{16} - 1 = \frac{25}{16} - \frac{16}{16} = \frac{9}{16}
\]
Now, multiplying both sides by 9:
\[
y^2 = 9 \cdot \frac{9}{16} = \frac{81}{16}
\]
Taking the square root gives:
\[
y = \frac{9}{4}
\]
Thus, point P is \( (5, \frac{9}{4}) \).
### Step 2: Find the equation of the tangent at point P
The formula for the tangent to the hyperbola at point \( (x_1, y_1) \) is given by:
\[
\frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1
\]
Here, \( a^2 = 16 \) and \( b^2 = 9 \), and \( (x_1, y_1) = (5, \frac{9}{4}) \).
Substituting these values into the tangent equation:
\[
\frac{5x}{16} - \frac{9y/4}{9} = 1
\]
This simplifies to:
\[
\frac{5x}{16} - \frac{y}{4} = 1
\]
Multiplying through by 16 to eliminate the denominators:
\[
5x - 4y = 16
\]
### Step 3: Find the points Q and R where the tangent meets the lines
The lines given are:
1. \( 3x - 4y = 0 \) (Line 1)
2. \( 3x + 4y = 0 \) (Line 2)
#### Finding point Q (intersection of tangent and Line 1)
Substituting \( 4y = 3x \) into the tangent equation:
\[
5x - 3x = 16 \implies 2x = 16 \implies x = 8
\]
Now substituting \( x = 8 \) back into \( 3x - 4y = 0 \):
\[
3(8) - 4y = 0 \implies 24 - 4y = 0 \implies 4y = 24 \implies y = 6
\]
Thus, point Q is \( (8, 6) \).
#### Finding point R (intersection of tangent and Line 2)
Substituting \( 4y = -3x \) into the tangent equation:
\[
5x + 3x = 16 \implies 8x = 16 \implies x = 2
\]
Now substituting \( x = 2 \) back into \( 3x + 4y = 0 \):
\[
3(2) + 4y = 0 \implies 6 + 4y = 0 \implies 4y = -6 \implies y = -\frac{3}{2}
\]
Thus, point R is \( (2, -\frac{3}{2}) \).
### Step 4: Calculate the area of triangle OQR
The vertices of triangle OQR are:
- O(0, 0)
- Q(8, 6)
- R(2, -\frac{3}{2})
Using the formula for the area of a triangle given vertices \( (x_1, y_1), (x_2, y_2), (x_3, y_3) \):
\[
\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
\]
Substituting the coordinates:
\[
\text{Area} = \frac{1}{2} \left| 0(6 + \frac{3}{2}) + 8(-\frac{3}{2} - 0) + 2(0 - 6) \right|
\]
Calculating:
\[
= \frac{1}{2} \left| 0 - 12 + (-12) \right| = \frac{1}{2} \left| -24 \right| = \frac{24}{2} = 12
\]
Thus, the area of triangle OQR is \( 12 \) square units.
### Final Answer
The area of triangle OQR is \( \boxed{12} \) square units.
