Question

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 the area of $ABCD$ is

Answer 1

Explanation for the correct option:

Step 1. Draw the graphs of $xy=1,xy=-1$:

${\left(xy\right)}^2=1$

$\Rightarrow$$xy=\pm 1$

Here, $OP=OQ$

$\Rightarrow \sqrt{{\alpha }^2+\frac{1}{{\alpha }^2}}=\sqrt{{\beta }^2+\frac{1}{{\beta }^2}}$

$\Rightarrow$ ${\alpha }^2+\frac{1}{{\alpha }^2}={\beta }^2+\frac{1}{{\beta }^2}$

$\Rightarrow$ ${\alpha }^2-{\beta }^2=\frac{1}{{\beta }^2}-\frac{1}{{\alpha }^2}$

$\Rightarrow$ ${\alpha }^2-{\beta }^2=\frac{{\alpha }^2-{\beta }^2}{{\alpha }^2{\beta }^2}$

$\Rightarrow$ ${\alpha }^2{\beta }^2=1$ ……(1)

Step 2. Find the value of $\alpha$ and $\beta$:

$A\left(\frac{\alpha +\beta }{2},\frac{{\displaystyle \frac{1}{\alpha }}-{\displaystyle \frac{1}{\beta }}}{2}\right)$ …$\left[\because AisthemidpointofPQandlieonthecurve{x}^2{y}^2=1\right]$

$\Rightarrow {\left(\frac{\alpha +\beta }{2}\right)}^2\times \left(\frac{1}{4}\right){\left(\frac{1}{\alpha }-\frac{1}{\beta }\right)}^2=1$

$\Rightarrow$ $\frac{{\left(\alpha +\beta \right)}^2}{4\times 4}\times {\left(\frac{\alpha -\beta }{\alpha \beta }\right)}^2=1$

$\Rightarrow$ ${\left[\left(\alpha +\beta \right)\left(\alpha -\beta \right)\right]}^2=16{\alpha }^2{\beta }^2$

$\Rightarrow$ ${\left({\alpha }^2-{\beta }^2\right)}^2=16$

$\Rightarrow$ ${\alpha }^2-{\beta }^2=\pm 4$

$\Rightarrow$ ${\alpha }^2-\frac{1}{{\alpha }^2}=\pm 4$ …[From equation (1)]

Step 3. Let ${\alpha }^2=t$

$\Rightarrow$ $t-\frac{1}{t}=\pm 4$

$\Rightarrow t^2\pm 4t-1=0$

$\Rightarrow$ $t=\frac{\pm 4\pm \sqrt{16+4}}{2}$

$\Rightarrow$ $t=\frac{\pm 4\pm 2\sqrt{5}}{2}$

$\Rightarrow$ ${\alpha }^2=\pm 2\pm \sqrt{5}$

$\Rightarrow$ ${\alpha }^2=2+\sqrt{5,}\sqrt{5}-2$

$\Rightarrow$ ${\beta }^2=\frac{1}{2+\sqrt{5}},\frac{1}{\sqrt{5}-2}$

Now, $O{P}^2={\alpha }^2+\frac{1}{{\alpha }^2}$

$=2+\surd 5+\frac{1}{(2+\surd 5)}$

0r, $O{P}^2=–2+\surd 5+\frac{1}{(-2+\surd 5)}$

$=2\surd 5$

Step 4. Find the area of square:

Area $=4\times \frac{1}{2}\times OP\times OQ$

$=2\times O{P}^2$

$=4\sqrt{5}$

Hence, the area of the square is $\mathbf{4}\sqrt{\mathbf{5}}$