JEE PYQ: Hyperbola Question 2
Question 2 - 2021 (18 Mar Shift 1)
A square ABCD has all its vertices on the curve $x^2y^2 = 1$. The midpoints of its sides also lie on the same curve. Then, the square of area of ABCD is ______.
Show Answer
Answer: 80
Solution
The curve is $xy = \pm 1$. Vertices: $A(t_1, \frac{1}{t_1})$, $B(t_2, -\frac{1}{t_2})$, etc. with $t_1, t_2 > 0$. From midpoint conditions: $t_1^2 - t_2^2 = 4$ and $t_1t_2 = 1$. So $t_1^2 + t_2^2 = \sqrt{16+4} = 2\sqrt{5}$. $AB^2 = (t_1-t_2)^2 + (\frac{1}{t_1}+\frac{1}{t_2})^2 = 2(t_1^2 + \frac{1}{t_1^2}) = 4\sqrt{5}$. Area$^2 = (AB^2)^2 = 80$.
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