An ellipse has equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $a$ and $b$ are positive real constants. Show that $\frac{\text{d}y}{\text{d}x} = -\frac{b^2 x}{a^2 y}$ if $y \ne 0$. The point $P(a\cos\theta, b\sin\theta)$, where $0 < \theta < \frac{\pi}{2}$, lies on the ellipse. Show that the equation of the tangent to the ellipse at $P$ is $\frac{x}{a}\cos\theta + \frac{y}{b}\sin\theta = 1$. The tangent found in part (b) meets the $x$-axis and $y$-axis at points $R$ and $S$ respectively. Find the area of triangle $ORS$ in terms of $\theta$, $a$ and $b$, where $O$ is the origin. Explain why the minimum area of triangle $ORS$ is $ab$ and state the value of $\theta$ which gives the minimum area of triangle $ORS$.

1. Differentiate both sides of the equation implicitly with respect to $x$.

$\frac{\text{d}}{\text{d}x}\left(\frac{x^2}{a^2} + \frac{y^2}{b^2}\right) = \frac{\text{d}}{\text{d}x}(1)$

$\frac{2x}{a^2} + \frac{2y}{b^2}\frac{\text{d}y}{\text{d}x} = 0$

2. Make $\frac{\text{d}y}{\text{d}x}$ the subject of the formula.

$\frac{2y}{b^2}\frac{\text{d}y}{\text{d}x} = -\frac{2x}{a^2}$

$\frac{\text{d}y}{\text{d}x} = -\frac{2x}{a^2} \times \frac{b^2}{2y}$

$\frac{\text{d}y}{\text{d}x} = -\frac{b^2 x}{a^2 y} \text{ (Shown)}$

1. Find the gradient of the tangent at point $P$.

From part (a), the gradient of the curve is given by $\frac{\text{d}y}{\text{d}x} = -\frac{b^2 x}{a^2 y}$.
Substitute $x = a\cos\theta$ and $y = b\sin\theta$:
$m = -\frac{b^2 (a\cos\theta)}{a^2 (b\sin\theta)} = -\frac{b\cos\theta}{a\sin\theta}$

2. Set up the equation of the tangent at $P$.

Using the point-gradient form $y - y_1 = m(x - x_1)$:
$y - b\sin\theta = -\frac{b\cos\theta}{a\sin\theta}(x - a\cos\theta)$

3. Rearrange to show the required equation.

Multiply both sides by $\frac{\sin\theta}{b}$:
$\frac{y}{b}\sin\theta - \sin^2\theta = -\frac{\cos\theta}{a}(x - a\cos\theta)$
$\frac{y}{b}\sin\theta - \sin^2\theta = -\frac{x}{a}\cos\theta + \cos^2\theta$
$\frac{x}{a}\cos\theta + \frac{y}{b}\sin\theta = \sin^2\theta + \cos^2\theta$
Since $\sin^2\theta + \cos^2\theta = 1$:
$\frac{x}{a}\cos\theta + \frac{y}{b}\sin\theta = 1 \quad \text{(shown)}$

1. Find the $x$-coordinate of point $R$ where the tangent meets the $x$-axis.

Substitute $y = 0$ into the equation of the tangent from part (b):
$\frac{x}{a}\cos\theta + 0 = 1$
$x = \frac{a}{\cos\theta}$
Since $0 < \theta < \frac{\pi}{2}$, $\cos\theta > 0$, so $OR = \frac{a}{\cos\theta}$.

2. Find the $y$-coordinate of point $S$ where the tangent meets the $y$-axis.

Substitute $x = 0$ into the equation of the tangent:
$0 + \frac{y}{b}\sin\theta = 1$
$y = \frac{b}{\sin\theta}$
Since $0 < \theta < \frac{\pi}{2}$, $\sin\theta > 0$, so $OS = \frac{b}{\sin\theta}$.

3. Calculate the area of the right-angled triangle $ORS$ and simplify.

Since $O$ is the origin, $\triangle ORS$ is a right-angled triangle with base $OR$ and height $OS$.
$\text{Area} = \frac{1}{2} \times OR \times OS$
$= \frac{1}{2} \left( \frac{a}{\cos\theta} \right) \left( \frac{b}{\sin\theta} \right)$
$= \frac{ab}{2\sin\theta\cos\theta}$
Using the double angle identity $\sin 2\theta = 2\sin\theta\cos\theta$:
$\text{Area} = \frac{ab}{\sin 2\theta}$

1. State the area of triangle $ORS$ using the coordinates from part (c)

From part (c), the coordinates of the intercepts $R$ and $S$ are $\left(\frac{a}{\cos\theta}, 0\right)$ and $\left(0, \frac{b}{\sin\theta}\right)$ respectively.

$\text{Area} = \frac{1}{2} \left(\frac{a}{\cos\theta}\right) \left(\frac{b}{\sin\theta}\right) = \frac{ab}{2\sin\theta\cos\theta} = \frac{ab}{\sin 2\theta}$

2. Explain the condition for the minimum area

For the area function $\frac{ab}{\sin 2\theta}$ to be a minimum, the denominator $\sin 2\theta$ must be at its maximum.

Since $0 < \theta < \frac{\pi}{2}$, the angle $2\theta$ is in the range $0 < 2\theta < \pi$. The maximum possible value of the sine function in this interval is $1$.

Therefore, the minimum area is:
$\frac{ab}{1} = ab$

3. Find the value of $\theta$ that gives the minimum area

The maximum value $\sin 2\theta = 1$ occurs when:
$2\theta = \frac{\pi}{2}$
$\theta = \frac{\pi}{4}$