Point O is the centre of the ellipse with maj | vedclass

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(A) Let the semi-major axis be $a$ and semi-minor axis be $b$. The coordinates are $O(0,0)$,$F(ae, 0)$,and $C(0, b)$. Given $OF = ae = 6$,so $a^2e^2 =

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$65$

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(A) Let the semi-major axis be $a$ and semi-minor axis be $b$. The coordinates are $O(0,0)$,$F(ae, 0)$,and $C(0, b)$. Given $OF = ae = 6$,so $a^2e^2 = 36$. Since $b^2 = a^2(1-e^2)$,we have $a^2 - b^2 = a^2e^2 = 36$ ... $(1)$.In $\Delta OCF$,the sides are $OC = b$,$OF = 6$,and $CF = \sqrt{b^2 + 36} = a$. The inradius $r$ of a right-angled triangle is given by $r = \frac{OC + OF - CF}{2}$.Given the diameter is $2$,so $r = 1$. Thus,$1 = \frac{b + 6 - a}{2}$ $\Rightarrow b + 6 - a = 2$ $\Rightarrow a - b = 4$ ... $(2)$.From $(1)$,$(a-b)(a+b) = 36$. Substituting $(2)$,$4(a+b) = 36 \Rightarrow a+b = 9$ ... $(3)$.Adding $(2)$ and $(3)$,$2a = 13 \Rightarrow a = 6.5$. Subtracting,$2b = 5 \Rightarrow b = 2.5$.The major axis $AB = 2a = 13$ and minor axis $CD = 2b = 5$.The product $(AB)(CD) = 13 	imes 5 = 65$.

Point $O$ is the centre of the ellipse with major axis $AB$ and minor axis $CD$. Point $F$ is one focus of the ellipse. If $OF = 6$ and the diameter of the inscribed circle of triangle $OCF$ is $2$,then the product $(AB)(CD)$ is equal to

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