A square ABCD has all its vertices on the curve x ^ 2 y ^ 2 = 1 ....

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

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

Correct answer

Step-by-step solution

$xy = 1, - 1$

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

$\dfrac{{t}_{1} + {t}_{2}}{2} \cdot \dfrac{\dfrac{1}{{t}_{1}} - \dfrac{1}{{t}_{2}}}{2} = 1$

$⇒ {t}_{2}^{2} - {t}_{1} = 2 4{t}_{1}{t}_{2}$

Product of slope $= - 1$

$1{t}_{1}^{2} \times \left(- 1{t}_{2}^{2}\right) = - 1 ⇒ {t}_{1}{t}_{2} = 1$

$⇒ {\left({t}_{1}{t}_{2}\right)}^{2} = 1 ⇒ {t}_{1}{t}_{2} = 1$

$⇒ {t}_{1}^{2} + {t}_{2}^{2} = \sqrt{{4}^{2} + 4} = 2\sqrt{5}$

$⇒ {t}_{1}^{2} = 2 + \sqrt{5} ⇒ 1{t}_{1}^{2} = \sqrt{5} - 2$

$A{B}^{2} = {\left({t}_{1} - {t}_{2}\right)}^{2} + {\left(\dfrac{1}{{t}_{1}} + \dfrac{1}{{t}_{2}}\right)}^{2}$

$= 2\left({t}_{1}^{2} + 1{t}_{1}^{2}\right) = 4\sqrt{5} ⇒ {\text{Area}}^{2} = 80$ sq. unit

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