Question
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:
Options
Solution
Sol.(a)
$$\begin{array}{r} a^{2} = b^{2} + 6^{2}\#(1) \end{array}$$
$$\begin{matrix} r & & \ = \frac{\Delta}{S} = \frac{\frac{1}{2} \times 6 \times b}{\frac{1}{2}(a + b + 6)} \\ & & \ = \frac{6b}{a + b + 6} = 1 \\ \Rightarrow 5b & \ = a + 6 & \text{(2)} \end{matrix}$$
From eqn. (1) and (2)
$$\begin{matrix} & & (5b - 6)^{2} & = b^{2} + 36 \\ \Rightarrow & 24b^{2} - 60b & & = 0 \\ \Rightarrow & b & & = \frac{5}{2} \\ & & a = \frac{25}{2} - 6 & = \frac{13}{2} \end{matrix}$$
$$(AB)(CD) = (2a)(2b) = 4 \times \frac{13}{2} \times \frac{5}{2} = 65$$
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