To find the area of triangle PQR formed by the tangent and normal to the ellipse \(3x^2 + 5y^2 = 32\) at the point \(P(2, 2)\), we will follow these steps:
### Step 1: Write the equation of the ellipse in standard form.
The given equation of the ellipse is:
\[
3x^2 + 5y^2 = 32
\]
To convert this to standard form, we divide the entire equation by 32:
\[
\frac{x^2}{\frac{32}{3}} + \frac{y^2}{\frac{32}{5}} = 1
\]
This gives us:
\[
\frac{x^2}{\frac{32}{3}} + \frac{y^2}{\frac{32}{5}} = 1
\]
From this, we identify:
\[
a^2 = \frac{32}{3}, \quad b^2 = \frac{32}{5}
\]
### Step 2: Find the equation of the tangent at point \(P(2, 2)\).
The equation of the tangent to the ellipse at point \((x_1, y_1)\) is given by:
\[
\frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1
\]
Substituting \(x_1 = 2\), \(y_1 = 2\), \(a^2 = \frac{32}{3}\), and \(b^2 = \frac{32}{5}\):
\[
\frac{2x}{\frac{32}{3}} + \frac{2y}{\frac{32}{5}} = 1
\]
Multiplying through by \( \frac{32}{15} \) to eliminate the denominators:
\[
3x + 5y = 16
\]
### Step 3: Find the point \(Q\) where the tangent meets the x-axis.
To find the x-intercept (point \(Q\)), set \(y = 0\) in the tangent equation:
\[
3x + 5(0) = 16 \implies 3x = 16 \implies x = \frac{16}{3}
\]
Thus, the coordinates of \(Q\) are:
\[
Q\left(\frac{16}{3}, 0\right)
\]
### Step 4: Find the equation of the normal at point \(P(2, 2)\).
The slope of the tangent is given by the equation \(3x + 5y = 16\). Rearranging gives:
\[
y = -\frac{3}{5}x + \frac{16}{5}
\]
Thus, the slope of the normal is the negative reciprocal:
\[
\text{slope of normal} = \frac{5}{3}
\]
Using the point-slope form of the equation of the normal:
\[
y - 2 = \frac{5}{3}(x - 2)
\]
Expanding this:
\[
y - 2 = \frac{5}{3}x - \frac{10}{3}
\]
Rearranging gives:
\[
5x - 3y + 4 = 0
\]
### Step 5: Find the point \(R\) where the normal meets the x-axis.
To find the x-intercept (point \(R\)), set \(y = 0\):
\[
5x - 3(0) + 4 = 0 \implies 5x + 4 = 0 \implies x = -\frac{4}{5}
\]
Thus, the coordinates of \(R\) are:
\[
R\left(-\frac{4}{5}, 0\right)
\]
### Step 6: Calculate the area of triangle \(PQR\).
The area \(A\) of triangle \(PQR\) can be calculated using the formula:
\[
A = \frac{1}{2} \times \text{base} \times \text{height}
\]
Here, the base is the distance between points \(Q\) and \(R\):
\[
\text{base} = \left|\frac{16}{3} - \left(-\frac{4}{5}\right)\right| = \left|\frac{16}{3} + \frac{4}{5}\right|
\]
Finding a common denominator (15):
\[
\frac{16}{3} = \frac{80}{15}, \quad -\frac{4}{5} = -\frac{12}{15}
\]
Thus,
\[
\text{base} = \frac{80 + 12}{15} = \frac{92}{15}
\]
The height is the y-coordinate of point \(P\), which is \(2\).
Now substituting into the area formula:
\[
A = \frac{1}{2} \times \frac{92}{15} \times 2 = \frac{92}{15}
\]
### Final Answer
The area of triangle \(PQR\) is:
\[
\frac{92}{15} \text{ square units.}
\]
