星形线（astroid）是具有参数方程
$x = a \cos^3 t, \ y = a \sin^3 t.$
的曲线。

1. Determine the axial intercepts by evaluating the parametric equations at the endpoints of the given interval.

When $t = 0$:
$x = 2 \cos^3(0) = 2$
$y = 2 \sin^3(0) = 0$
The $x$-intercept is $(2, 0)$.

When $t = \frac{\pi}{2}$:
$x = 2 \cos^3\left(\frac{\pi}{2}\right) = 0$
$y = 2 \sin^3\left(\frac{\pi}{2}\right) = 2$
The $y$-intercept is $(0, 2)$.

2. Sketch the curve for the given domain, ensuring correct shape and bounds.

For $0 < t < \frac{\pi}{2}$, both $x > 0$ and $y > 0$.
Using a graphing calculator or considering the powers, the curve connects $(2, 0)$ and $(0, 2)$ entirely in the first quadrant. It is concave upwards (bowing inwards towards the origin).
(The sketch must show a smooth curve starting at $(2,0)$ with a horizontal tangent and ending at $(0,2)$ with a vertical tangent, clearly labelled with the coordinates $(2,0)$ and $(0,2)$.)

1. Find the derivatives of $x$ and $y$ with respect to $t$.

Given $a = 1$, we have $x = \cos^3 t$ and $y = \sin^3 t$.
Using the chain rule:
$\frac{dx}{dt} = 3\cos^2 t (-\sin t) = -3\sin t \cos^2 t$
$\frac{dy}{dt} = 3\sin^2 t (\cos t) = 3\sin^2 t \cos t$

2. Determine the gradient of the curve at $t = p$.

Using the parametric differentiation formula:
$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{3\sin^2 t \cos t}{-3\sin t \cos^2 t} = -\frac{\sin t}{\cos t} = -\tan t$
At point $P$ where $t = p$, the gradient of the tangent, $m$, is $-\tan p$.

3. Form the equation of the tangent at $P$.

At $t = p$, the coordinates of point $P$ are $(\cos^3 p, \sin^3 p)$.
Using the point-gradient equation of a straight line, $y - y_1 = m(x - x_1)$:
$y - \sin^3 p = -\tan p \, (x - \cos^3 p)$

4. Simplify the equation to reach the required form.

Express $\tan p$ as $\frac{\sin p}{\cos p}$ for the constant term:
$y = (-\tan p)x + (\tan p)(\cos^3 p) + \sin^3 p$
$y = (-\tan p)x + \left(\frac{\sin p}{\cos p}\right)(\cos^3 p) + \sin^3 p$
$y = (-\tan p)x + \sin p \cos^2 p + \sin^3 p$
Factorise $\sin p$ from the last two terms:
$y = (-\tan p)x + \sin p (\cos^2 p + \sin^2 p)$
Using the identity $\cos^2 p + \sin^2 p = 1$:
$y = (-\tan p)x + \sin p \quad (\text{shown})$

1. Find the coordinates of the intercepts $Q$ and $R$.

To find $Q$ (the $x$-intercept), substitute $y=0$ into the tangent equation $y = (-\tan p)x + \sin p$:
$0 = (-\tan p)x + \sin p$
$\left(\frac{\sin p}{\cos p}\right)x = \sin p \implies x = \cos p$
So, $Q$ has coordinates $(\cos p, 0)$.

To find $R$ (the $y$-intercept), substitute $x=0$ into the tangent equation:
$y = (-\tan p)(0) + \sin p \implies y = \sin p$
So, $R$ has coordinates $(0, \sin p)$.

2. Express the coordinates of the midpoint of $QR$ in terms of $p$.

Let the midpoint of $QR$ be $(x, y)$. Using the midpoint formula:
$x = \frac{\cos p + 0}{2} = \frac{\cos p}{2}$
$y = \frac{0 + \sin p}{2} = \frac{\sin p}{2}$
These represent the parametric equations for the path traced by the midpoint.

3. Eliminate the parameter $p$ to obtain the cartesian equation.

Rearranging the parametric equations gives:
$\cos p = 2x$
$\sin p = 2y$
Using the standard trigonometric identity $\cos^2 p + \sin^2 p = 1$, we substitute the expressions for $\cos p$ and $\sin p$:
$(2x)^2 + (2y)^2 = 1$
$4x^2 + 4y^2 = 1$
Dividing by 4 yields the equation of a circle:
$x^2 + y^2 = \frac{1}{4}$