The diagram shows a parabola formed by the intersection of a right circular cone and a plane parallel to the slanted edge of the cone. The cone with the point O at the centre of its base has radius $R$ cm and slant length $L$ cm, where $R$ and $L$ are constants. The points A, B, C and D on the plane are such that $BD = h$ cm, $AB = BC = x$ cm and the area $S$ of the region bounded by the parabola and the line $AC$ is $\frac{4}{3} x h$.

1. Find the distance from the centre of the base to the chord $AC$.

Let $O$ be the centre of the circular base. Due to the symmetry of the right circular cone and the plane being parallel to a slanted edge, the chord $AC$ on the base is perpendicular to the diameter passing through its midpoint $B$.
Triangle $OBC$ is a right-angled triangle with hypotenuse $OC = R$ (the radius of the base) and $BC = x$.
By Pythagoras' theorem,
$OB = \sqrt{OC^2 - BC^2} = \sqrt{R^2 - x^2}$

2. Express $h$ in terms of $R$, $L$, and $x$ using similar triangles.

Consider the vertical cross-section of the cone through the diameter passing through $B$. Let the diameter be $PQ$, with $O$ as the midpoint and $B$ located between the left endpoint $P$ and $O$. Let $V$ be the vertex of the cone.
The cross-section is $\triangle VPQ$. The slant edges are $VP = VQ = L$, and the diameter $PQ = 2R$.
Because the cutting plane is parallel to the slanted edge $VP$, the segment $BD$ on the plane is parallel to $VP$.
Therefore, $\triangle QBD$ is similar to $\triangle QPV$.
Using the ratio of corresponding sides:
$\frac{BD}{VP} = \frac{QB}{QP}$
Based on the diagram, $B$ is on the same side of $O$ as the parallel slanted edge, so the distance from the opposite endpoint $Q$ to $B$ is:
$QB = QO + OB = R + \sqrt{R^2 - x^2}$
Substitute $BD = h$, $VP = L$, and $QP = 2R$ into the ratio:
$\frac{h}{L} = \frac{R + \sqrt{R^2 - x^2}}{2R}$
$h = \frac{L\left(R + \sqrt{R^2 - x^2}\right)}{2R}$

3. Substitute $h$ to show the final expression for $S$.

Given that the area of the parabolic region is $S = \frac{4}{3}xh$, substitute the expression obtained for $h$:
$S = \frac{4}{3}x \left[ \frac{L\left(R + \sqrt{R^2 - x^2}\right)}{2R} \right]$
$S = \frac{4Lx\left(R + \sqrt{R^2 - x^2}\right)}{6R}$
$S = \frac{2Lx\left(R + \sqrt{R^2 - x^2}\right)}{3R} \quad \text{(shown)}$

1. Differentiate the given expression for $S$ with respect to $x$.

Given $S = \frac{2L}{3R} \left( Rx + x\sqrt{R^2 - x^2} \right)$,
$\frac{\mathrm{d}S}{\mathrm{d}x} = \frac{2L}{3R} \left( R + \sqrt{R^2 - x^2} + x \left[ \frac{1}{2}(R^2 - x^2)^{-\frac{1}{2}} (-2x) \right] \right)$
$\frac{\mathrm{d}S}{\mathrm{d}x} = \frac{2L}{3R} \left( R + \sqrt{R^2 - x^2} - \frac{x^2}{\sqrt{R^2 - x^2}} \right)$

2. Equate the derivative to zero to find the maximum point and rearrange the equation.

For maximum $S$, set $\frac{\mathrm{d}S}{\mathrm{d}x} = 0$:
$R + \sqrt{R^2 - x^2} - \frac{x^2}{\sqrt{R^2 - x^2}} = 0$
Multiply through by $\sqrt{R^2 - x^2}$:
$R\sqrt{R^2 - x^2} + (R^2 - x^2) - x^2 = 0$
$R\sqrt{R^2 - x^2} = 2x^2 - R^2$

3. Solve the algebraic equation for $x$ in terms of $R$.

Squaring both sides:
$R^2(R^2 - x^2) = (2x^2 - R^2)^2$
$R^4 - R^2 x^2 = 4x^4 - 4R^2 x^2 + R^4$
$4x^4 - 3R^2 x^2 = 0$
Since $x > 0$ (as it represents a physical length), we divide by $x^2$:
$4x^2 = 3R^2 \implies x = \frac{\sqrt{3}}{2}R$

4. Determine $h$ at the maximum point and calculate the ratio $\frac{h}{x}$.

From $S = \frac{4}{3} x h$, divide by $\frac{4}{3}x$:
$h = \frac{L}{2 R} \left( R + \sqrt{R^2 - x^2} \right)$
Substituting $x = \frac{\sqrt{3}}{2}R$ into the expression for $h$:
$h = \frac{L}{2 R} \left( R + \sqrt{R^2 - \frac{3}{4}R^2} \right) = \frac{L}{2 R} \left( R + \frac{1}{2}R \right) = \frac{3L}{4}$
Finally, compute $\frac{h}{x}$:
$\frac{h}{x} = \frac{\frac{3}{4}L}{\frac{\sqrt{3}}{2}R} = \frac{3L}{4} \times \frac{2}{\sqrt{3}R} = \frac{3L}{2\sqrt{3}R} = \frac{\sqrt{3}L}{2R}$