MCQ
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
- A$70$
- B$96$
- C$75$
- ✓$80$
✓
Answer
Correct option: D.
$80$
d
$xy =1,-1$
$xy =1,-1$
$\frac{t_{1}+t_{2}}{2} \cdot \frac{\frac{1}{t_{1}}-\frac{1}{t_{2}}}{2}=1$
$\Rightarrow t_{1}^{2}-t_{2}^{2}=4 t_{1} t_{2}$
$\frac{1}{t_{1}^{2}} \times\left(-\frac{1}{t_{2}^{2}}\right)=-1 \Rightarrow t_{1} t_{2}=1$
$\Rightarrow\left(t_{1} t_{2}\right)^{2}=1 \Rightarrow t_{1} t_{2}=1$
$t_{1}^{2}-t_{2}^{2}=4$
$\Rightarrow t_{1}^{2}+t_{2}^{2}=\sqrt{4^{2}+4}=2 \sqrt{5}$
$A B^{2}=\left(t_{1}-t_{2}\right)^{2}+\left(\frac{1}{t_{1}}+\frac{1}{t_{2}}\right)^{2}$
$\Rightarrow t_{1}^{2}=2+\sqrt{5} \Rightarrow \frac{1}{t_{1}^{2}}=\sqrt{5}-2$
$=2\left(t_{1}^{2}+\frac{1}{t_{1}^{2}}\right)=4 \sqrt{5} \Rightarrow$ Area $^{2}=80$
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