Rectangular hyperbola xy = c² — properties and parametric form

Question

For the rectangular hyperbola $xy = c^2$ , find its parametric form, the equation of the tangent at the point $\left(ct, \dfrac{c}{t}\right)$ , and the equation of the normal. Also find the condition for four points on this hyperbola to be concyclic.

(JEE Advanced 2021, similar pattern)

Solution — Step by Step

Any point on $xy = c^2$ can be written as:

P = \left(ct,\ \frac{c}{t}\right), \quad t \neq 0

Verify: $x \cdot y = ct \cdot \frac{c}{t} = c^2$ . Works for every non-zero $t$ .

Differentiate $xy = c^2$ implicitly: $y + x\frac{dy}{dx} = 0$ , so $\frac{dy}{dx} = -\frac{y}{x}$ .

At $P$ : slope $= -\frac{c/t}{ct} = -\frac{1}{t^2}$ .

Tangent: $y - \frac{c}{t} = -\frac{1}{t^2}(x - ct)$

yt^2 - ct = -x + ct

\mathbf{x + yt^2 = 2ct}

Normal is perpendicular to the tangent, so its slope is $t^2$ .

y - \frac{c}{t} = t^2(x - ct)

\mathbf{xt^3 - yt = ct^4 - c}

Or equivalently: $t^3 x - ty - c(t^4 - 1) = 0$ .

Four points $P(ct_i, c/t_i)$ for $i = 1, 2, 3, 4$ lie on a circle if and only if:

\mathbf{t_1 \cdot t_2 \cdot t_3 \cdot t_4 = 1}

This is a classic result. It comes from substituting the parametric points into the general circle equation $x^2 + y^2 + 2gx + 2fy + k = 0$ and using the fact that the resulting equation in $t$ has product of roots = 1.

Why This Works

The rectangular hyperbola $xy = c^2$ is special because its asymptotes ( $x$ -axis and $y$ -axis) are perpendicular — hence “rectangular.” It is actually the standard hyperbola $X^2 - Y^2 = c^2$ rotated by 45°.

The parametric form $\left(ct, c/t\right)$ is elegant because it uses a single parameter and every algebraic property becomes a property of $t$ . The concyclic condition $t_1 t_2 t_3 t_4 = 1$ is used heavily in JEE Advanced problems on conics.

Key properties: eccentricity $e = \sqrt{2}$ (always), and the asymptotes are the coordinate axes themselves.

Alternative Method

For the tangent equation, you can also use the formula: the tangent at $(x_1, y_1)$ on $xy = c^2$ is $xy_1 + x_1 y = 2c^2$ . This is the " $T = 0$ " form. Substituting $(x_1, y_1) = (ct, c/t)$ : $x \cdot \frac{c}{t} + ct \cdot y = 2c^2 \Rightarrow x + yt^2 = 2ct$ . Same result.

For JEE Advanced, the rectangular hyperbola is tested every 2-3 years. The most common question type: given three points on $xy = c^2$ with parameters $t_1, t_2, t_3$ , find the fourth point such that all four are concyclic. Use $t_4 = 1/(t_1 t_2 t_3)$ .

Common Mistake

Students often confuse the tangent formula for $xy = c^2$ with the tangent formula for $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ . These are different conics with different formulas. For the rectangular hyperbola, the tangent at $(ct, c/t)$ is $x + yt^2 = 2ct$ — not the standard conic tangent formula. Keep the two separate in your formula sheet.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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