Key Concepts and Formulas

This problem requires us to find the area of a triangle formed by a point on an ellipse and the x-intercepts of the tangent and normal lines drawn to the ellipse at that point. We will utilize the following key concepts and formulas:

1. Equation of Tangent to an Ellipse: For an ellipse given by $Ax^2 + By^2 = C$, the equation of the tangent at a point $P(x_1, y_1)$ lying on the ellipse is given by the $T=0$ form: $Ax x_1 + By y_1 = C$
2. Equation of Normal to an Ellipse: The normal line to a curve at a point is perpendicular to the tangent line at that same point. If the slope of the tangent is $m_T$, then the slope of the normal $m_N$ is its negative reciprocal, i.e., $m_N = -\frac{1}{m_T}$ (provided $m_T \neq 0$). The equation of the normal can then be found using the point-slope form: $y - y_1 = m_N (x - x_1)$
3. X-intercept: The x-intercept of a line is the point where the line crosses the x-axis. At this point, the y-coordinate is 0.
4. Area of a Triangle: For a triangle with vertices $P(x_P, y_P)$, $Q(x_Q, 0)$, and $R(x_R, 0)$, where the base QR lies on the x-axis, the area is given by: $\text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} |x_Q - x_R| |y_P|$

Given Information:

• Equation of the ellipse: $3x^2 + 5y^2 = 32$
• Point P on the ellipse: $P(2, 2)$
• Q and R are the x-intercepts of the tangent and normal at P, respectively.

Our goal is to find the area of triangle PQR.

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Step 1: Verify the point P lies on the ellipse

Why this step? Before proceeding with tangent and normal calculations, it's good practice to ensure the given point actually lies on the curve. If it doesn't, the question might imply a different type of tangent (from an external point) or be flawed.

Substitute the coordinates of $P(2, 2)$ into the equation of the ellipse: $3(2)^2 + 5(2)^2 = 3(4) + 5(4) = 12 + 20 = 32$ Since $32 = 32$, the point $P(2, 2)$ indeed lies on the ellipse.

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Step 2: Find the equation of the tangent at P(2,2) and its x-intercept Q

Why this step? We need the tangent line to find its x-intercept, which is point Q. The $T=0$ form provides a direct and efficient way to find the tangent equation for conics.

Using the $T=0$ form for the ellipse $3x^2 + 5y^2 = 32$ at $P(x_1, y_1) = (2, 2)$: $3x x_1 + 5y y_1 = 32$ Substitute $x_1 = 2$ and $y_1 = 2$: $3x(2) + 5y(2) = 32$ $6x + 10y = 32$ Divide the entire equation by 2 to simplify: $3x + 5y = 16$ This is the equation of the tangent line.

To find the x-intercept Q, we set $y = 0$ in the tangent equation: $3x + 5(0) = 16$ $3x = 16$ $x = \frac{16}{3}$ So, the x-intercept of the tangent is $Q\left(\frac{16}{3}, 0\right)$.

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Step 3: Find the equation of the normal at P(2,2) and its x-intercept R

Why this step? We need the normal line to find its x-intercept, which is point R. To find the normal, we first need the slope of the tangent, as the normal is perpendicular to it.

From the tangent equation $3x + 5y = 16$, we can find its slope $m_T$. Rewrite it in slope-intercept form $y = mx + c$: $5y = -3x + 16$ $y = -\frac{3}{5}x + \frac{16}{5}$ So, the slope of the tangent is $m_T = -\frac{3}{5}$.

The normal line is perpendicular to the tangent line. Therefore, the slope of the normal $m_N$ is the negative reciprocal of $m_T$: $m_N = -\frac{1}{m_T} = -\frac{1}{(-3/5)} = \frac{5}{3}$

Now, use the point-slope form $y - y_1 = m_N (x - x_1)$ with $P(x_1, y_1) = (2, 2)$ and $m_N = \frac{5}{3}$ to find the equation of the normal: $y - 2 = \frac{5}{3}(x - 2)$ Multiply by 3 to clear the denominator: $3(y - 2) = 5(x - 2)$ $3y - 6 = 5x - 10$ Rearrange the terms to get the standard form: $5x - 3y - 4 = 0$ This is the equation of the normal line.

To find the x-intercept R, we set $y = 0$ in the normal equation: $5x - 3(0) - 4 = 0$ $5x - 4 = 0$ $5x = 4$ $x = \frac{4}{5}$ So, the x-intercept of the normal is $R\left(\frac{4}{5}, 0\right)$.

Common Mistake Alert: Be careful with the signs when calculating the negative reciprocal of the slope. A common error is to forget the negative sign or invert it incorrectly.

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Step 4: Calculate the area of triangle PQR

Why this step? We have the coordinates of all three vertices P, Q, and R. Since Q and R lie on the x-axis, the segment QR forms the base of the triangle. This makes the area calculation straightforward using the base-height formula.

The coordinates of the vertices are:

• $P(2, 2)$
• $Q\left(\frac{16}{3}, 0\right)$
• $R\left(\frac{4}{5}, 0\right)$

The base of the triangle, QR, lies on the x-axis. Its length is the absolute difference of the x-coordinates of Q and R: $\text{Base} = |x_Q - x_R| = \left|\frac{16}{3} - \frac{4}{5}\right|$ To subtract these fractions, find a common denominator (which is 15): $\text{Base} = \left|\frac{16 \times 5}{3 \times 5} - \frac{4 \times 3}{5 \times 3}\right| = \left|\frac{80}{15} - \frac{12}{15}\right| = \left|\frac{80 - 12}{15}\right| = \left|\frac{68}{15}\right| = \frac{68}{15}$

The height of the triangle is the perpendicular distance from P to the x-axis, which is the absolute value of the y-coordinate of P: $\text{Height} = |y_P| = |2| = 2$

Now, calculate the area of triangle PQR: $\text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height}$ $\text{Area} = \frac{1}{2} \times \frac{68}{15} \times 2$ $\text{Area} = \frac{68}{15}$

Thus, the area of triangle PQR is $\frac{68}{15}$ square units.

Tip: When one side of a triangle lies on an axis, using the $\frac{1}{2} \times \text{base} \times \text{height}$ formula is generally much simpler and less prone to calculation errors than using the determinant formula for the area of a triangle.

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Summary and Key Takeaway

In this problem, we systematically found the area of triangle PQR by:

1. Confirming the point P lies on the ellipse.
2. Determining the equation of the tangent at P using the $T=0$ form and finding its x-intercept Q.
3. Determining the equation of the normal at P by first finding the tangent's slope, then its negative reciprocal for the normal's slope, and finding its x-intercept R.
4. Finally, calculating the area of triangle PQR using the base-height formula, leveraging the fact that the base QR lies on the x-axis.

The area of triangle PQR is $\frac{68}{15}$ square units.

The final answer is $\boxed{\text{C}}$.