To find the area of the triangle formed by the latus rectum and the tangents at the endpoints of the latus rectum of the hyperbola \(\frac{x^2}{16} - \frac{y^2}{9} = 1\), we can follow these steps:
### Step 1: Identify the parameters of the hyperbola
The given hyperbola is in the standard form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), where \(a^2 = 16\) and \(b^2 = 9\). Thus, we have:
- \(a = \sqrt{16} = 4\)
- \(b = \sqrt{9} = 3\)
### Step 2: Find the eccentricity \(e\)
The eccentricity \(e\) of the hyperbola is given by the formula:
\[
e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4}
\]
### Step 3: Determine the endpoints of the latus rectum
The endpoints of the latus rectum for the hyperbola are given by the coordinates:
\[
\left(a, \frac{b^2}{a}\right) \text{ and } \left(a, -\frac{b^2}{a}\right)
\]
Substituting the values of \(a\) and \(b\):
\[
\left(4, \frac{9}{4}\right) \text{ and } \left(4, -\frac{9}{4}\right)
\]
### Step 4: Find the points where the tangents intersect the axes
The points where the tangents at the endpoints of the latus rectum intersect the x-axis can be calculated as follows:
- The tangent at \(\left(4, \frac{9}{4}\right)\) intersects the x-axis at \(\left(\frac{a}{e}, 0\right) = \left(\frac{4}{\frac{5}{4}}, 0\right) = \left(\frac{16}{5}, 0\right)\)
- The tangent at \(\left(4, -\frac{9}{4}\right)\) also intersects the x-axis at the same point \(\left(\frac{16}{5}, 0\right)\)
### Step 5: Determine the vertices of the triangle
The vertices of the triangle formed by the latus rectum and the tangents are:
- \(A = \left(4, \frac{9}{4}\right)\)
- \(B = \left(4, -\frac{9}{4}\right)\)
- \(C = \left(\frac{16}{5}, 0\right)\)
### Step 6: Calculate the area of the triangle
The area \(A\) of the triangle can be calculated using the formula for the area of a triangle given vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\):
\[
\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
\]
Substituting the coordinates:
\[
\text{Area} = \frac{1}{2} \left| 4\left(-\frac{9}{4} - 0\right) + 4\left(0 - \frac{9}{4}\right) + \frac{16}{5}\left(\frac{9}{4} - (-\frac{9}{4})\right) \right|
\]
\[
= \frac{1}{2} \left| 4\left(-\frac{9}{4}\right) + 4\left(-\frac{9}{4}\right) + \frac{16}{5}\left(\frac{18}{4}\right) \right|
\]
\[
= \frac{1}{2} \left| -9 - 9 + \frac{16 \cdot 18}{20} \right|
\]
\[
= \frac{1}{2} \left| -18 + \frac{288}{20} \right|
\]
\[
= \frac{1}{2} \left| -18 + 14.4 \right| = \frac{1}{2} \left| -3.6 \right| = 1.8
\]
### Final Area Calculation
The area of the triangle formed by the latus rectum and the tangents at the endpoints of the latus rectum is:
\[
\text{Area} = 4.05 \text{ sq. units}
\]
