A square ABCD has all its vertices on the curve x²y² = 1. The midpoints of its sides also lie on the

Q: Wait, why do we assume the square is centered at the origin? Couldn't it be shifted?

That is a fantastic question! The curve $x^2y^2 = 1$ is perfectly symmetric about the origin. If you were to shift the square away from the origin, you would break the symmetry required for all four vertices to lie on these specific hyperbolic branches. In geometry, when a polygon is inscribed in a symmetric curve, its center of symmetry must coincide with the curve's center of symmetry. Trust the symmetry!

Q: Why did we take the positive root when solving $(t^2 - 1/t^2)^2 = 16$?

Great catch! We have $(t^2 - 1/t^2)^2 = 16$, which means $t^2 - 1/t^2$ could be $4$ or $-4$. However, we placed vertex $A$ in the first quadrant where $t > 0$. By our construction, $A$ is $(t, 1/t)$ and $B$ is $(-1/t, t)$. For the square to exist in the first quadrant as we visualized, $t$ must be greater than $1/t$, making $t^2 - 1/t^2$ positive. If we chose $-4$, it would imply $t < 1/t$, which is just a reflection of the same geometry.

Q: Why can't we just use the distance formula directly to find the area without all this algebra?

You absolutely can! The distance formula gives you the side length $AB$. The area is $AB^2$. The reason we use the algebraic identity is to avoid solving for $t$ explicitly. Solving for $t$ would involve a quartic equation, which is messy and prone to errors. By using the identity $(t^2 + 1/t^2)^2 = (t^2 - 1/t^2)^2 + 4$, we bypass the need to find $t$ and jump straight to the area. It's the elegant, 'JEE-style' way to solve it!

Question

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

Sigma Percentile · Accepted Answer

### The Geometry of Symmetry

Welcome, future engineer! Today, we are going to tackle a problem that might look like a nightmare of algebra, but is actually a beautiful dance of symmetry. We are dealing with a square $ABCD$ whose vertices and midpoints all lie on the curve $x^2y^2 = 1$.

### Phase 1: The Battlefield

First, let's look at the curve $x^2y^2 = 1$. If we take the square root of both sides, we get $xy = 1$ or $xy = -1$. These are classic rectangular hyperbolas.

They form four symmetric branches across the four quadrants. Because the curve is symmetric about the origin $(0,0)$, any square inscribed within it must also be centered at the origin.

If it were shifted, the vertices would lose their perfect alignment with the hyperbolic branches. This is our first major insight: the center of our square is $(0,0)$.

### Phase 2: Defining the Vertices

Let's place vertex $A$ on the branch $xy = 1$ in the first quadrant. We can define its coordinates parametrically as $A\left(t, \frac{1}{t}ight)$, where $t > 0$.

Now, since the square is centered at the origin, vertex $B$ is simply a $90^\circ$ rotation of $A$. Using the rotation rule for coordinate geometry, rotating $(x, y)$ by $90^\circ$ counterclockwise gives us $(-y, x)$.

Applying this to $A$, we get $B\left(-\frac{1}{t}, tight)$. If you check, you will see that $B$ lies on the branch $xy = -1$, which is exactly where it should be!

### Phase 3: The Midpoint Constraint

Now, the problem tells us that the midpoints of the sides also lie on the curve. Let's find the midpoint $M$ of side $AB$. Using the midpoint formula, we get:

$$M = \left( \frac{t - \frac{1}{t}}{2}, \frac{t + \frac{1}{t}}{2} ight)$$

Since $M$ lies on $x^2y^2 = 1$, its coordinates must satisfy the equation. Substituting these into $x^2y^2 = 1$, we get:

$$\left( \frac{t - \frac{1}{t}}{2} ight)^2 \left( \frac{t + \frac{1}{t}}{2} ight)^2 = 1$$

This looks intimidating, but watch the magic happen. The numerator becomes $\left(t^2 - \frac{1}{t^2}ight)^2$ and the denominator becomes $16$. So:

$$\left( t^2 - \frac{1}{t^2} ight)^2 = 16 \Rightarrow t^2 - \frac{1}{t^2} = 4$$

### Phase 4: The Final Calculation

We need the area of the square, $S = AB^2$. Using the distance formula between $A\left(t, \frac{1}{t}ight)$ and $B\left(-\frac{1}{t}, tight)$, we find:

$$S = \left( t + \frac{1}{t} ight)^2 + \left( \frac{1}{t} - t ight)^2$$

Expanding this, we get $S = 2\left(t^2 + \frac{1}{t^2}ight)$. We know $t^2 - \frac{1}{t^2} = 4$, but we need $t^2 + \frac{1}{t^2}$.

We use the identity $\left(t^2 + \frac{1}{t^2}ight)^2 = \left(t^2 - \frac{1}{t^2}ight)^2 + 4$. Substituting $4$, we get $16 + 4 = 20$.

Thus, $t^2 + \frac{1}{t^2} = \sqrt{20}$. The area $S = 2\sqrt{20}$.

The question asks for the square of the area, $S^2 = (2\sqrt{20})^2 = 4 	imes 20 = \mathbf{80}$. And there you have it! A beautiful, logical path to the answer.

Animated Solution for Mathematics - Conic Sections: 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

Enter Numerical Value:

Visualized Solution

Analyze the Curve $x^{2} y^{2} = 1$

Symmetry and Square Placement

Defining Vertex $A (t, \frac{1}{t})$

Finding Vertex $B$ via Rotation

Locating the Midpoint $M$

Applying the Curve Condition

Simplifying the Expression

Solving for $t^{2} - \frac{1}{t ^{2}}$

Calculating the Area of the Square

Expanding the Area Expression

Using the Algebraic Identity

Finding the Square of the Area

The Sigma Insight: Rectangular Hyperbola

The Geometry of Symmetry

Phase 1

Phase 2

Phase 3

Phase 4

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