Since point \( Q(x, y) \) lies on the circle \( x^2 + y^2 = 9 \):

\[ y^2 = 9 – x^2 \]

Since \( Q \) is in the first quadrant (\( y > 0 \)):

\[ y = \sqrt{9 – x^2} \]

Answer: \( y = \sqrt{9 – x^2} \)

Find the area \( A \) of triangle \( PQR \) with vertices \( P(-3, 0) \), \( Q(x, \sqrt{9 – x^2}) \), \( R(x, -\sqrt{9 – x^2}) \).

Method 1: Determinant Formula

Use the formula for the area of a triangle with vertices \( (x_1, y_1) \), \( (x_2, y_2) \), \( (x_3, y_3) \):

\[ A = \frac{1}{2} \left| x_1 (y_2 – y_3) + x_2 (y_3 – y_1) + x_3 (y_1 – y_2) \right| \]

Substitute coordinates:

\[ A = \frac{1}{2} \left| (-3) (\sqrt{9 – x^2} – (-\sqrt{9 – x^2})) + x (-\sqrt{9 – x^2} – 0) + x (0 – \sqrt{9 – x^2}) \right| \]

\[ = \frac{1}{2} \left| (-3) (2\sqrt{9 – x^2}) + x (-\sqrt{9 – x^2}) + x (-\sqrt{9 – x^2}) \right| \]

\[ = \frac{1}{2} \left| -6\sqrt{9 – x^2} – 2x \sqrt{9 – x^2} \right| = \frac{1}{2} \left| (-6 – 2x) \sqrt{9 – x^2} \right| \]

Since \( x \geq 0 \), \( -6 – 2x < 0 \), so:

\[ A = \frac{1}{2} (6 + 2x) \sqrt{9 – x^2} = (3 + x) \sqrt{9 – x^2} \]

Method 2: Base and Height

Base \( QR \): Distance between \( Q(x, \sqrt{9 – x^2}) \) and \( R(x, -\sqrt{9 – x^2}) \):

\[ QR = \sqrt{(x – x)^2 + (\sqrt{9 – x^2} – (-\sqrt{9 – x^2}))^2} = \sqrt{(2\sqrt{9 – x^2})^2} = 2\sqrt{9 – x^2} \]

Height: Perpendicular distance from \( P(-3, 0) \) to line \( QR \). Line \( QR \) is horizontal at \( x = x \), so height is the x-distance from \( x = -3 \) to \( x \):

\[ h = x – (-3) = x + 3 \]

\[ A = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot 2\sqrt{9 – x^2} \cdot (x + 3) = (3 + x) \sqrt{9 – x^2} \]

Answer: \( A = (3 + x) \sqrt{9 – x^2} \)

Show \( \frac{dA}{dx} = \frac{9 – 3x – 2x^2}{\sqrt{9 – x^2}} \).

Method 1: Product Rule

For \( A = (3 + x) \sqrt{9 – x^2} \), let \( u = 3 + x \), \( v = \sqrt{9 – x^2} \).

\[ \frac{du}{dx} = 1, \quad \frac{dv}{dx} = \frac{d}{dx} (9 – x^2)^{1/2} = \frac{1}{2} (9 – x^2)^{-1/2} (-2x) = -\frac{x}{\sqrt{9 – x^2}} \]

Apply product rule: \( \frac{dA}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \):

\[ \frac{dA}{dx} = (3 + x) \left( -\frac{x}{\sqrt{9 – x^2}} \right) + \sqrt{9 – x^2} \cdot 1 \]

\[ = -\frac{x (3 + x)}{\sqrt{9 – x^2}} + \sqrt{9 – x^2} \]

Combine over common denominator:

\[ = \frac{-x (3 + x) + (9 – x^2)}{\sqrt{9 – x^2}} = \frac{-3x – x^2 + 9 – x^2}{\sqrt{9 – x^2}} = \frac{9 – 3x – 2x^2}{\sqrt{9 – x^2}} \]

Method 2: Chain Rule with Substitution

Express \( A = (3 + x) y \), where \( y = \sqrt{9 – x^2} \), so \( \frac{dy}{dx} = -\frac{x}{\sqrt{9 – x^2}} \).

\[ \frac{dA}{dx} = \frac{dA}{dy} \cdot \frac{dy}{dx} \]

\[ \frac{dA}{dy} = x + 3, \quad \frac{dy}{dx} = -\frac{x}{\sqrt{9 – x^2}} \]

\[ \frac{dA}{dx} = (x + 3) \left( -\frac{x}{\sqrt{9 – x^2}} \right) + \sqrt{9 – x^2} \cdot 1 = \frac{-x (x + 3) + (9 – x^2)}{\sqrt{9 – x^2}} \]

\[ = \frac{9 – 3x – 2x^2}{\sqrt{9 – x^2}} \]

Answer: \( \frac{dA}{dx} = \frac{9 – 3x – 2x^2}{\sqrt{9 – x^2}} \)

Find the y-coordinate of \( R \) that maximizes \( A \). Set \( \frac{dA}{dx} = 0 \):

\[ \frac{9 – 3x – 2x^2}{\sqrt{9 – x^2}} = 0 \]

Since \( \sqrt{9 – x^2} \neq 0 \) for \( x \in (-3, 3) \), solve:

\[ 9 – 3x – 2x^2 = 0 \implies 2x^2 + 3x – 9 = 0 \]

Quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \), with \( a = 2 \), \( b = 3 \), \( c = -9 \):

\[ x = \frac{-3 \pm \sqrt{3^2 – 4 \cdot 2 \cdot (-9)}}{2 \cdot 2} = \frac{-3 \pm \sqrt{9 + 72}}{4} = \frac{-3 \pm \sqrt{81}}{4} = \frac{-3 \pm 9}{4} \]

\[ x = \frac{6}{4} = \frac{3}{2}, \quad x = \frac{-12}{4} = -3 \]

Since \( Q(x, y) \) is in the first quadrant (\( x > 0 \)), take \( x = \frac{3}{2} \).

Find \( y \) for \( Q \):

\[ y = \sqrt{9 – \left(\frac{3}{2}\right)^2} = \sqrt{9 – \frac{9}{4}} = \sqrt{\frac{36}{4} – \frac{9}{4}} = \sqrt{\frac{27}{4}} = \frac{\sqrt{27}}{2} = \frac{3\sqrt{3}}{2} \]

Since \( R(x, -y) \), the y-coordinate of \( R \) is:

\[ y_R = -\frac{3\sqrt{3}}{2} \]

Verify maximum using second derivative (optional):

\[ \frac{dA}{dx} = \frac{9 – 3x – 2x^2}{\sqrt{9 – x^2}} \]

Use quotient rule for \( \frac{d^2 A}{dx^2} \), but first derivative test suffices per markscheme. At \( x = \frac{3}{2} \), \( \frac{dA}{dx} = 0 \). Test points: For \( x < \frac{3}{2} \), e.g., \( x = 1 \), numerator \( 9 – 3 \cdot 1 – 2 \cdot 1^2 = 4 > 0 \); for \( x > \frac{3}{2} \), e.g., \( x = 2 \), \( 9 – 3 \cdot 2 – 2 \cdot 2^2 = -5 < 0 \). Thus, \( x = \frac{3}{2} \) is a maximum.

Answer: y-coordinate of \( R \) is \( -\frac{3\sqrt{3}}{2} \)