Obtain a tangent-intercept triangle and qualify its endpoint

$y=(1-x)^{3/2}$ for $0\le x\le1$. The curve joins the included endpoints $(1,0)$ and $(0,1)$, decreasing and convex as a graph of $x$. As $t$ increases, it runs from $(1,0)$ to $(0,1)$.

$dx/dt=-2\cos t\sin t$ and $dy/dt=3\sin^2t\cos t$. In the interior, $dy/dx=-3\sin t/2$, so the tangent is $y-\sin^3\theta=-\tfrac32\sin\theta(x-\cos^2\theta)$. The one-sided Cartesian derivative gives the same expression at both endpoints, although both parameter derivatives vanish there.

For $0<\theta\le\pi/2$, putting $y=0$ gives $Q=(\cos^2\theta+\tfrac23\sin^2\theta,0)$. Putting $x=0$ gives $R=(0,\tfrac32\sin\theta\cos^2\theta+\sin^3\theta)$. Write $A=3\cos^2\theta+2\sin^2\theta$; the positive intercept lengths are $OQ=A/3$ and $OR=(\sin\theta)A/2$.

The area is $\tfrac12(A/3)[(\sin\theta)A/2]=\tfrac1{12}\sin\theta(3\cos^2\theta+2\sin^2\theta)^2$, as required. At $\theta=0$, the tangent is the x-axis itself, so its x-axis intersection $Q$ is not unique and a nondegenerate triangle is not defined. The printed formula has limit zero there; it is interpreted as the degenerate limiting area, not as a unique pair of tangent intercepts.