In short

The conjugate hyperbola of \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 is \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1 — the same asymptotes, but the branches open along the other axis. When a = b, the hyperbola becomes rectangular: its asymptotes are perpendicular, its eccentricity is exactly \sqrt{2}, and after a 45° rotation the equation simplifies to the clean form xy = c^2.

Take the hyperbola \frac{x^2}{9} - \frac{y^2}{4} = 1. Its branches open left and right, hugging two diagonal lines called asymptotes. Now ask: what curve opens up and down, has the exact same asymptotes, and fits into the gaps left behind by the original hyperbola?

That curve is \frac{y^2}{4} - \frac{x^2}{9} = 1. Swap the sign. That is all. The branches that were horizontal become vertical. The asymptotes stay identical. The two hyperbolas together tile the plane around the origin like four wings of a pinwheel, each pair of branches nestled in the gaps left by the other.

This twin is called the conjugate hyperbola. Studying it is not an academic exercise — the relationship between a hyperbola and its conjugate reveals structural properties that you cannot see from one curve alone. And the special case where the two curves are the same shape — the rectangular hyperbola — turns out to be one of the most important curves in coordinate geometry.

The conjugate hyperbola

Start with the standard hyperbola

H: \quad \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1

This curve has its transverse axis along the x-axis, vertices at (\pm a, 0), and branches opening left and right. Its asymptotes are the lines y = \pm \frac{b}{a}x.

Conjugate hyperbola

The conjugate hyperbola of \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 is

H': \quad \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1

equivalently written as -\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, or \frac{x^2}{a^2} - \frac{y^2}{b^2} = -1.

The conjugate hyperbola has its transverse axis along the y-axis. Its vertices are at (0, \pm b) and its branches open up and down. The roles of a and b flip: the transverse axis of H' has semi-length b, and the conjugate axis of H' has semi-length a.

The hyperbola $\frac{x^2}{9} - \frac{y^2}{4} = 1$ (black, opening left-right) and its conjugate $\frac{y^2}{4} - \frac{x^2}{9} = 1$ (red, opening up-down). Both share the same asymptotes $y = \pm \frac{2}{3}x$ (dashed). The four branches fill the four regions created by the asymptotes.

Relationship between a hyperbola and its conjugate

The hyperbola and its conjugate are deeply linked. Here is a side-by-side comparison.

Property Hyperbola H Conjugate H'
Equation \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1
Transverse axis Along x-axis, length 2a Along y-axis, length 2b
Conjugate axis Along y-axis, length 2b Along x-axis, length 2a
Vertices (\pm a, 0) (0, \pm b)
Foci (\pm c, 0) where c^2 = a^2 + b^2 (0, \pm c') where c'^2 = a^2 + b^2
Eccentricity e = c/a e' = c'/b
Asymptotes y = \pm \frac{b}{a}x y = \pm \frac{b}{a}x

Several things are worth noticing.

Same asymptotes. Both H and H' share the exact same pair of asymptotes y = \pm \frac{b}{a}x. This is the defining relationship. The asymptotes are the skeleton that both hyperbolas are built on — one opens between the acute angles of the asymptotes, the other opens between the obtuse angles.

Same value of c. Since c^2 = a^2 + b^2 for both curves, the distance from the centre to each focus is the same number c for both the hyperbola and its conjugate. The foci of H lie on the x-axis; the foci of H' lie on the y-axis; but all four foci sit on a circle of radius c centred at the origin.

Different eccentricities (in general). The eccentricity of H is e = c/a, and the eccentricity of H' is e' = c/b. These are equal only when a = b — a case worth examining closely.

The sum identity. Add the equations of H and H':

\frac{x^2}{a^2} - \frac{y^2}{b^2} + \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1 + 1

The left side is 0 and the right side is 2, so 0 = 2 — which is nonsense. This tells you that no point lies on both H and H'. The two curves never intersect.

The difference identity. Subtract instead:

\frac{x^2}{a^2} - \frac{y^2}{b^2} - \left(\frac{y^2}{b^2} - \frac{x^2}{a^2}\right) = 1 - 1
\frac{2x^2}{a^2} - \frac{2y^2}{b^2} = 0 \implies \frac{x^2}{a^2} = \frac{y^2}{b^2} \implies y = \pm \frac{b}{a}x

Those are the asymptotes. So the asymptotes are literally the "average" of the hyperbola and its conjugate — the curve you get when you set the two equations equal.

Eccentricity relation

There is a clean identity linking the eccentricities. Let e = c/a and e' = c/b where c^2 = a^2 + b^2.

\frac{1}{e^2} + \frac{1}{e'^2} = \frac{a^2}{c^2} + \frac{b^2}{c^2} = \frac{a^2 + b^2}{c^2} = \frac{c^2}{c^2} = 1

So the eccentricities of a hyperbola and its conjugate always satisfy

\frac{1}{e^2} + \frac{1}{e'^2} = 1

This is a striking identity. It says that knowing one eccentricity pins down the other. If e = 2, then \frac{1}{4} + \frac{1}{e'^2} = 1, so e' = \frac{2}{\sqrt{3}}.

All four foci of the hyperbola $\frac{x^2}{9} - \frac{y^2}{4} = 1$ and its conjugate lie on the circle $x^2 + y^2 = 13$ (dotted), since $c^2 = 9 + 4 = 13$ for both. The hyperbola's foci sit on the $x$-axis at $(\pm\sqrt{13}, 0)$; the conjugate's foci sit on the $y$-axis at $(0, \pm\sqrt{13})$. Four foci, one circle.

Rectangular hyperbola: the case a = b

When a = b, the hyperbola \frac{x^2}{a^2} - \frac{y^2}{a^2} = 1 simplifies to

x^2 - y^2 = a^2

This is called a rectangular hyperbola. The name comes from a geometric fact: its asymptotes are perpendicular.

The asymptotes of a general hyperbola are y = \pm \frac{b}{a}x. When a = b, these become y = x and y = -x. The slopes are +1 and -1. The product of these slopes is (-1), which means the asymptotes are perpendicular. Two perpendicular lines form a rectangle (with the axes), hence the name.

Rectangular hyperbola

A rectangular hyperbola (also called an equilateral hyperbola) is a hyperbola with a = b. In standard position:

x^2 - y^2 = a^2

Its asymptotes y = x and y = -x are perpendicular. Its eccentricity is exactly \sqrt{2}.

The eccentricity is always \sqrt{2}

For a rectangular hyperbola, a = b, so c^2 = a^2 + b^2 = 2a^2, which gives c = a\sqrt{2}. The eccentricity is

e = \frac{c}{a} = \frac{a\sqrt{2}}{a} = \sqrt{2}

This is a fixed number — it does not depend on the size of the hyperbola. Every rectangular hyperbola, regardless of how large or small, has eccentricity \sqrt{2}. Among all hyperbolas, the rectangular hyperbola is the one closest to being "circular" in spirit (eccentricity as close to 1 as a hyperbola can get while still being a hyperbola, since e > 1 for all hyperbolas).

You can also verify the eccentricity identity: \frac{1}{e^2} + \frac{1}{e'^2} = \frac{1}{2} + \frac{1}{2} = 1. When the hyperbola is rectangular, its conjugate is also rectangular, and both have the same eccentricity \sqrt{2}.

The rectangular hyperbola $x^2 - y^2 = 4$ (black) and its conjugate $y^2 - x^2 = 4$ (red), with $a = b = 2$. The asymptotes $y = x$ and $y = -x$ are perpendicular. The foci of the original hyperbola lie at $(\pm 2\sqrt{2}, 0) \approx (\pm 2.83, 0)$, and the foci of the conjugate lie at $(0, \pm 2\sqrt{2})$. All four foci sit on a circle of radius $2\sqrt{2}$.

The xy = c^2 form

A rectangular hyperbola x^2 - y^2 = a^2 has perpendicular asymptotes y = x and y = -x. Now imagine rotating the entire figure by 45° so that the asymptotes align with the coordinate axes. What does the equation become?

The rotation

To rotate the coordinate system by 45° anticlockwise, substitute

x = \frac{X - Y}{\sqrt{2}}, \qquad y = \frac{X + Y}{\sqrt{2}}

where (X, Y) are the new coordinates. Substitute into x^2 - y^2 = a^2:

\left(\frac{X - Y}{\sqrt{2}}\right)^2 - \left(\frac{X + Y}{\sqrt{2}}\right)^2 = a^2
\frac{X^2 - 2XY + Y^2}{2} - \frac{X^2 + 2XY + Y^2}{2} = a^2
\frac{-4XY}{2} = a^2
-2XY = a^2
XY = -\frac{a^2}{2}

Writing c^2 = \frac{a^2}{2}, this becomes XY = -c^2. If you had instead rotated by 45° clockwise (or equivalently, taken the conjugate hyperbola y^2 - x^2 = a^2), you would get XY = c^2 with a positive sign.

Either way, the equation is of the form xy = k for some constant k. In most textbooks, the standard form is written with the positive constant:

Rectangular hyperbola in rotated form

A rectangular hyperbola with asymptotes along the coordinate axes has the equation

xy = c^2

where c > 0. The asymptotes are the x-axis (y = 0) and the y-axis (x = 0). The curve lives in the first and third quadrants. The transverse axis lies along y = x, and the vertices are at (c, c) and (-c, -c).

The beauty of this form is its simplicity. The function y = c^2/x is just a scaled version of y = 1/x. Every time you have graphed y = k/x in class, you were graphing a rectangular hyperbola.

Key properties of xy = c^2

Vertices. The vertices are the points closest to the centre (origin). On xy = c^2, the transverse axis is along y = x. Setting y = x gives x^2 = c^2, so x = \pm c. The vertices are (c, c) and (-c, -c).

Eccentricity. Since xy = c^2 is a rectangular hyperbola, its eccentricity is \sqrt{2}, regardless of the value of c.

Foci. The foci lie on the transverse axis y = x, at distance a_{\text{rot}}\sqrt{2} from the centre, where a_{\text{rot}} = c\sqrt{2} is the semi-transverse axis length. So the foci are at (c\sqrt{2}, c\sqrt{2}) and (-c\sqrt{2}, -c\sqrt{2}) — but these calculations are rarely needed in practice.

Parametric form. The curve xy = c^2 has an elegant parametric representation:

x = ct, \qquad y = \frac{c}{t}, \qquad t \neq 0

For any non-zero t, the point (ct, c/t) lies on xy = c^2 because (ct)(c/t) = c^2. As t varies over all non-zero reals, the point traces the entire curve.

The rectangular hyperbola $xy = 4$ (so $c = 2$). The asymptotes are the coordinate axes themselves. The transverse axis is $y = x$ (dashed), and the vertices are $(2, 2)$ and $(-2, -2)$. The conjugate axis $y = -x$ (dotted) passes through the centre. Notice the symmetry: the point $(1, 4)$ and the point $(4, 1)$ are reflections in the line $y = x$.

Why xy = c^2 matters

The form xy = c^2 appears naturally in many settings:

Worked examples

Example 1: Find the conjugate hyperbola and its eccentricity

Given the hyperbola \frac{x^2}{16} - \frac{y^2}{9} = 1, find its conjugate hyperbola, the eccentricities of both, and verify the identity \frac{1}{e^2} + \frac{1}{e'^2} = 1.

Step 1. Identify a and b.

a^2 = 16, \; b^2 = 9 \implies a = 4, \; b = 3

Why: compare with the standard form \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1.

Step 2. Write the conjugate hyperbola.

\frac{y^2}{9} - \frac{x^2}{16} = 1

Why: swap the sign — replace the 1 on the right with -1, or equivalently write \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1.

Step 3. Find c and the eccentricities.

c^2 = a^2 + b^2 = 16 + 9 = 25, \quad c = 5
e = \frac{c}{a} = \frac{5}{4}, \qquad e' = \frac{c}{b} = \frac{5}{3}

Why: for the original hyperbola the transverse axis has semi-length a = 4, so e = c/a. For the conjugate, the transverse axis has semi-length b = 3, so e' = c/b.

Step 4. Verify the identity.

\frac{1}{e^2} + \frac{1}{e'^2} = \frac{1}{(5/4)^2} + \frac{1}{(5/3)^2} = \frac{16}{25} + \frac{9}{25} = \frac{25}{25} = 1 \checkmark

Why: this always works because a^2 + b^2 = c^2. The identity is just Pythagoras in disguise.

Result: The conjugate hyperbola is \frac{y^2}{9} - \frac{x^2}{16} = 1, with e = \frac{5}{4} and e' = \frac{5}{3}. The identity holds.

The hyperbola $\frac{x^2}{16} - \frac{y^2}{9} = 1$ (black) and its conjugate $\frac{y^2}{9} - \frac{x^2}{16} = 1$ (red). The foci of both lie on a circle of radius $c = 5$ centred at the origin: $(\pm 5, 0)$ for the original, $(0, \pm 5)$ for the conjugate.

The picture confirms: both curves share the same asymptotes, and the foci of the original at (\pm 5, 0) and the foci of the conjugate at (0, \pm 5) are all at distance 5 from the centre.

Example 2: Tangent to a rectangular hyperbola in $xy = c^2$ form

A point P on the rectangular hyperbola xy = 9 has x-coordinate 1. Find the equation of the tangent at P.

Step 1. Find the coordinates of P.

xy = 9, \; x = 1 \implies y = 9

So P = (1, 9).

Why: substitute the known x-coordinate into the curve equation to get y.

Step 2. Differentiate xy = 9 implicitly to find the slope.

\frac{d}{dx}(xy) = 0 \implies y + x\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{y}{x}

Why: the product rule gives \frac{d}{dx}(xy) = y \cdot 1 + x \cdot \frac{dy}{dx}, and the right side is constant so its derivative is 0.

Step 3. Evaluate the slope at P = (1, 9).

\frac{dy}{dx}\bigg|_{(1,9)} = -\frac{9}{1} = -9

Why: substitute the coordinates of P into the slope formula.

Step 4. Write the tangent equation using point-slope form.

y - 9 = -9(x - 1)
y = -9x + 9 + 9 = -9x + 18

Why: the tangent line passes through P = (1, 9) with slope -9.

Result: The tangent at P = (1, 9) is y = -9x + 18, or equivalently 9x + y = 18.

The rectangular hyperbola $xy = 9$ and its tangent at $P = (1, 9)$. The tangent has slope $-9$ and meets the $x$-axis at $(2, 0)$. The steep tangent reflects the fact that near $x = 1$ the curve $y = 9/x$ is dropping rapidly.

Notice something about the intercepts of the tangent: it meets the x-axis at (2, 0) and the y-axis at (0, 18). The midpoint of these intercepts is (1, 9) — exactly the point P. This is not a coincidence: for any tangent to xy = c^2, the point of tangency is always the midpoint of the segment cut off by the axes. This elegant property is unique to rectangular hyperbolas.

Common confusions

Going deeper

If you came here for the conjugate hyperbola and the xy = c^2 form, you have them. The rest of this section explores the tangent-intercept property and a connection to the area of a triangle formed by any tangent to a rectangular hyperbola.

The tangent-intercept property of xy = c^2

Take any point P = (ct, c/t) on the rectangular hyperbola xy = c^2. The tangent at P can be derived using the parametric form.

Since x = ct and y = c/t:

\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-c/t^2}{c} = -\frac{1}{t^2}

The tangent at P = (ct, c/t) is:

y - \frac{c}{t} = -\frac{1}{t^2}(x - ct)
yt^2 - ct = -x + ct
x + yt^2 = 2ct

This line meets the x-axis (set y = 0) at x = 2ct, and the y-axis (set x = 0) at y = 2c/t.

The x-axis intercept is A = (2ct, 0) and the y-axis intercept is B = (0, 2c/t).

Midpoint of AB: \left(\frac{2ct}{2}, \frac{2c/t}{2}\right) = (ct, c/t) = P.

So P is always the midpoint of the intercept segment, regardless of where P sits on the curve. This is a beautiful invariant.

Constant area of the tangent triangle

The triangle OAB — formed by the origin and the two intercepts — has area

\text{Area} = \frac{1}{2} |OA| \cdot |OB| = \frac{1}{2} \cdot 2ct \cdot \frac{2c}{t} = 2c^2

The t cancels. The area is 2c^2, a constant independent of P. No matter where you draw the tangent on xy = c^2, the triangle it forms with the axes always has the same area.

This constant-area property is one of the reasons the rectangular hyperbola is a favourite in JEE problems — it gives you a fixed quantity to work with, and many problems reduce to using this fact.

Connection to Bhaskara II's work

The rectangular hyperbola appears implicitly in Bhaskara II's Lilavati (12th century) through problems involving inverse proportion. When Bhaskara writes about quantities where "as one increases, the other decreases in the same ratio," the underlying curve is xy = k — a rectangular hyperbola. The explicit identification of this as a conic section came later, but the curve itself was already present in Indian mathematical thought through practical problems of commerce and astronomy.

Where this leads next