The diagram shows a parabola formed by the intersection of a right circular cone and a plane paralle

Question

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 $\frac{4}{3}xh$. Show that $S=\frac{2Lx\left( R+\sqrt{{{R}^{2}}-{{x}^{2}}} \right)}{3R}$. [2] It is given that $S$ is a maximum. Use the differentiation to find $\frac{h}{x}$ in terms of $R$ and $L$. [5]

Accepted Answer

(i) $S = \frac{2Lx(R + \sqrt{R^2 - x^2})}{3R}$ (ii) $\frac{h}{x} = \frac{\sqrt{3}L}{2R}$

2020 YIJC P1 Q5

2020 YIJC P1 Q5

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