A square ABCD has all its vertices on the cur | vedclass

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Question

$xy =1,-1$

$\frac{t_{1}+t_{2}}{2} \cdot \frac{\frac{1}{t_{1}}-\frac{1}{t_{2}}}{2}=1$

$\Rightarrow t_{1}^{2}-t_{2}^{2}=4 t_{1} t_{2}$

$\frac{1

Vedclass · Accepted Answer

$80$

Vedclass · Answer

$xy =1,-1$

$\frac{t_{1}+t_{2}}{2} \cdot \frac{\frac{1}{t_{1}}-\frac{1}{t_{2}}}{2}=1$

$\Rightarrow t_{1}^{2}-t_{2}^{2}=4 t_{1} t_{2}$

$\frac{1}{t_{1}^{2}} 	imes\left(-\frac{1}{t_{2}^{2}}ight)=-1 \Rightarrow t_{1} t_{2}=1$

$\Rightarrow\left(t_{1} t_{2}ight)^{2}=1 \Rightarrow t_{1} t_{2}=1$

$t_{1}^{2}-t_{2}^{2}=4$

$\Rightarrow t_{1}^{2}+t_{2}^{2}=\sqrt{4^{2}+4}=2 \sqrt{5}$

$A B^{2}=\left(t_{1}-t_{2}ight)^{2}+\left(\frac{1}{t_{1}}+\frac{1}{t_{2}}ight)^{2}$

$\Rightarrow t_{1}^{2}=2+\sqrt{5} \Rightarrow \frac{1}{t_{1}^{2}}=\sqrt{5}-2$

$=2\left(t_{1}^{2}+\frac{1}{t_{1}^{2}}ight)=4 \sqrt{5} \Rightarrow$ Area $^{2}=80$

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

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