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  • A square ABCD has all its vertices on the curve The midpoints of its sides also lie on the same curve. Then, the square of area of ABCD is ____________.   O

A square ABCD has all its vertices on the curve x^{2}y^{2}=1. The midpoints of its sides also lie on the same curve. Then, the square of area of ABCD is ____________.  
Option: 1 80
Option: 2 60
Option: 3 4
Option: 4 25

Answers (1)

best_answer

\\\mathrm{OA} \perp \mathrm{OB} \\ \Rightarrow \left ( \frac{0-\left (-\frac{1}{q} \right )}{0-q} \right )\left ( \frac{0-\frac{1}{p}}{0-p} \right )=-1 \\ \Rightarrow\left(\frac{1}{\mathrm{p}^{2}}\right)\left(-\frac{1}{\mathrm{q}^{2}}\right)=-1\\\Rightarrow x^2y^2=1

\\\Rightarrow(p+q)^{2}\left(\frac{1}{p}-\frac{1}{q}\right)^{2}=16 \\ \Rightarrow(p+q)^{2}(p-q)^{2}=16 \\ \Rightarrow\left(p^{2}-q^{2}\right)^{2}=16 \\ \Rightarrow p^{2}-\frac{1}{p^{2}}=\pm 4

\\\Rightarrow \mathrm{p}^{4} \pm 4 \mathrm{p}^{2}-1=0 \\ \Rightarrow \mathrm{p}^{2}=\frac{\pm 4 \pm \sqrt{20}}{2}=\pm 2 \pm \sqrt{5} \\ \Rightarrow \mathrm{p}^{2}=2+\sqrt{5} \text { or }-2+\sqrt{5}

\\\mathrm{OB}^{2}=\mathrm{p}^{2}+\frac{1}{\mathrm{p}^{2}}=2+\sqrt{5}+\frac{1}{2+\sqrt{5}} \text { or }-2+\sqrt{5}+\frac{1}{-2+\sqrt{5}}=2 \sqrt{5} \\ \text { Area }=4\left(\frac{1}{2}\right)(\mathrm{OA})(\mathrm{OB})=2(\mathrm{OB})^{2}=4\sqrt5

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himanshu.meshram

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