JEE MAIN - Mathematics (2021 - 18th March Morning Shift - No. 22)
A square ABCD has all its vertices on the curve x2y2 = 1. The midpoints of its sides also lie on the same curve. Then, the square of area of ABCD is _________.
Answer
80
Explanation
x2y2 = 1
$$ \Rightarrow $$ y2 = $${1 \over {{x^2}}}$$
$$ \Rightarrow $$ y = $$ \pm {1 \over x}$$
Graph of this equation,
$$OA \bot OB$$
$$ \Rightarrow \left( {{1 \over {{p^2}}}} \right)\left( { - {1 \over {{q^2}}}} \right) = - 1$$
$$ \Rightarrow {p^2}{q^2} = 1$$
$$P\left( {{{p + q} \over 2},{{{1 \over p} - {1 \over q}} \over 2}} \right)$$ midpoint of AB lies
On $${x^2}{y^2} = 1$$
$$ \Rightarrow {(p + q)^2}{\left( {{1 \over p} - {1 \over q}} \right)^2} = 16$$
$$ \Rightarrow {(p + q)^2}{(p - q)^2} = 16$$
$$ \Rightarrow {({p^2} - {q^2})^2} = 16$$
$$ \Rightarrow {P^2} - {1 \over {{P^2}}} = \pm 4$$
$$ \Rightarrow {p^4} \pm 4{p^2} - 1 = 0$$
$$ \Rightarrow {p^2} = {{ \pm 4 \pm \sqrt {20} } \over 2} = \pm 2 \pm \sqrt 5 $$
$$ \Rightarrow {p^2} = 2 + \sqrt 5 $$ or $$ - 2 + \sqrt 5 $$
$$O{B^2} = {p^2} + {1 \over {{p^2}}} = 2 + \sqrt 5 + {1 \over {2 + \sqrt 5 }}$$ or $$ - 2 + \sqrt 5 + {1 \over { - 2 + \sqrt 5 }} = 2\sqrt 5 $$
Area $$ = 4\left( {{1 \over 2}} \right)(OA)(OB) = 2{(OB)^2} = 4\sqrt 5 $$_18th_March_Morning_Shift_en_22_1.png)
$$ \Rightarrow $$ y2 = $${1 \over {{x^2}}}$$
$$ \Rightarrow $$ y = $$ \pm {1 \over x}$$
Graph of this equation,
$$OA \bot OB$$
$$ \Rightarrow \left( {{1 \over {{p^2}}}} \right)\left( { - {1 \over {{q^2}}}} \right) = - 1$$
$$ \Rightarrow {p^2}{q^2} = 1$$
$$P\left( {{{p + q} \over 2},{{{1 \over p} - {1 \over q}} \over 2}} \right)$$ midpoint of AB lies
On $${x^2}{y^2} = 1$$
$$ \Rightarrow {(p + q)^2}{\left( {{1 \over p} - {1 \over q}} \right)^2} = 16$$
$$ \Rightarrow {(p + q)^2}{(p - q)^2} = 16$$
$$ \Rightarrow {({p^2} - {q^2})^2} = 16$$
$$ \Rightarrow {P^2} - {1 \over {{P^2}}} = \pm 4$$
$$ \Rightarrow {p^4} \pm 4{p^2} - 1 = 0$$
$$ \Rightarrow {p^2} = {{ \pm 4 \pm \sqrt {20} } \over 2} = \pm 2 \pm \sqrt 5 $$
$$ \Rightarrow {p^2} = 2 + \sqrt 5 $$ or $$ - 2 + \sqrt 5 $$
$$O{B^2} = {p^2} + {1 \over {{p^2}}} = 2 + \sqrt 5 + {1 \over {2 + \sqrt 5 }}$$ or $$ - 2 + \sqrt 5 + {1 \over { - 2 + \sqrt 5 }} = 2\sqrt 5 $$
Area $$ = 4\left( {{1 \over 2}} \right)(OA)(OB) = 2{(OB)^2} = 4\sqrt 5 $$
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