Rectangular hyperbola xy = c²: Question bank

Definition The cast of symbols (all built here, so nothing is assumed)

$c$ is a positive fixed number with $x y = c^{2}$ . We take == $c > 0$ == throughout: the vertex (defined below) then reaches $(c, c)$ , and all sign arguments below rely on this choice. (If someone hands you a negative constant $k$ with $x y = k$ , rename $c^{2} = ∣ k ∣$ and adjust the branch — never carry a negative $c$ .)
$t$ is the parameter: one real number naming the point $(c t, \frac{c}{t})$ , with $t \neq = 0$ . Check it lies on the curve: $c t \cdot \frac{c}{t} = c^{2}$ . ✓ A pair of parameters $t_{1}, t_{2}$ (written with subscripts throughout: $t_{1}$ , $t_{2}$ , and their product $t_{1} t_{2}$ ) just names two such points — the endpoints of a chord.
The centre is the point where the two asymptotes cross: here the origin $(0, 0)$ . It is the curve's centre of symmetry — reflecting any curve point through it lands on another curve point. The curve itself does not pass through the centre.
The asymptotes are the two coordinate axes $x = 0$ and $y = 0$ — the lines the curve races toward but never touches. They cross at the centre.
A vertex is a point of the curve closest to the centre — the "tip" of a branch. On $x y = c^{2}$ the two vertices are $(c, c)$ and $(- c, - c)$ ; you can see they are the nearest curve points to the origin because $x^{2} + y^{2} = (c t)^{2} + (c / t)^{2}$ is smallest exactly at $t = 1$ (and $t = - 1$ ).
The transverse axis (the line joining the two vertices, through the centre) is $y = x$ ; the conjugate axis (perpendicular to it, through the centre) is $y = - x$ .
A standard hyperbola here means $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ , where $a$ is the distance from centre to vertex along the x-direction and $b$ is the matching semi-axis in the y-direction. "Rectangular" is the special case == $a = b$ ==.

Look at s01: the two asymptotes, the centre where they cross, the two vertices on $y = x$ , and the constant-area rectangle are all labelled on one board.

Definition What "rotate the axes" means here (built on this page, not deferred)

Several traps mention turning $x^{2} - y^{2} = a^{2}$ into $x y = c^{2}$ by rotating the axes by $4 5^{\circ}$ . Here is what that phrase means so nothing is left to an outside link.

Keep every point of the curve fixed; only turn the grid you measure with by $4 5^{\circ}$ . Call the new (turned) coordinates $X$ and $Y$ — so $X$ is the reading along the tilted horizontal axis and $Y$ along the tilted vertical axis. Same point, new labels.
The turning formulas (from $cos 4 5^{\circ} = sin 4 5^{\circ} = \frac{1}{2}$ ) are $x = \frac{X - Y}{2}$ , $y = \frac{X + Y}{2}$ .
Substitute into $x^{2} - y^{2}$ : $x^{2} - y^{2} = \frac{( X - Y ) ^{2} - ( X + Y ) ^{2}}{2} = \frac{- 4 X Y}{2} = - 2 X Y .$ The factor of 2 appears because $cos 4 5^{\circ} sin 4 5^{\circ} = \frac{1}{2}$ — that is the whole source of the $2$ . So $- 2 X Y = a^{2}$ , giving $X Y = - \frac{a ^{2}}{2}$ . Relabel the turned coordinates as ordinary $x, y$ and pick the branch so the product is positive: $x y = c^{2}$ with $c^{2} = \frac{a ^{2}}{2}$ . See also Rotation of axes for the general tool.

Formula The three formulas these traps refer to (mini-derivations, so they're earned)

Tangent at parameter $t$ . Differentiate $x y = c^{2}$ : $y + x \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = - \frac{y}{x}$ . At $(c t, c / t)$ the slope is $- \frac{c / t}{c t} = - \frac{1}{t ^{2}}$ . Point-slope gives $y - \frac{c}{t} = - \frac{1}{t ^{2}} (x - c t)$ . Now we tidy: multiplying both sides by $t^{2}$ and dividing by $t$ are reversible algebraic steps — they never add or drop solutions, so the line is exactly the same set of points, just written more symmetrically. The result: $\frac{x}{t} + t y = 2 c .$ Normal at $t$ . The normal is the line perpendicular to the tangent at the same point (it points "straight out" from the curve). Perpendicular lines have slopes that multiply to $- 1$ , so the normal slope is the negative reciprocal of $- 1/ t^{2}$ , namely $+ t^{2}$ . Point-slope then clears denominators (again a reversible step, so the line is unchanged), giving $x t^{3} - y t = c (t^{4} - 1)$ . Chord joining $t_{1}, t_{2}$ . Through $(c t_{1}, c / t_{1})$ and $(c t_{2}, c / t_{2})$ the slope is $\frac{c / t _{1} - c / t _{2}}{c t _{1} - c t _{2}} = - \frac{1}{t _{1} t _{2}}$ ; point-slope, cleared of fractions (reversible, so still the same line), tidies to $x + t_{1} t_{2} y = c (t_{1} + t_{2}) .$ Setting $t_{1} = t_{2} = t$ collapses it to the tangent — a built-in check.

s02 shows the point $(c t, c / t)$ , its tangent (chalk blue) and the perpendicular normal (chalk pink), so you can see why the slopes are negative-reciprocal.

True or false — justify

x y = c^{2}

is a brand-new curve, unrelated to

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

False. It is the same curve as

x^{2} - y^{2} = a^{2}

(the

a = b

standard hyperbola), just viewed with the axes rotated

4 5^{\circ}

onto the asymptotes; substituting the rotation gives

- 2 X Y = a^{2}

(where

X, Y

are the turned coordinates, as built in the rotation box), so

c^{2} = a^{2} /2

Every rectangular hyperbola has eccentricity

2

, no matter the value of

c

True. Eccentricity measures shape, not size;

c

only scales the curve. For

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

the eccentricity is

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

; with

a = b

this is

1 + 1 = 2

for every rectangular hyperbola (see the derivation box above).

The coordinate axes are the transverse and conjugate axes of

x y = c^{2}

False. For

x y = c^{2}

the coordinate axes are the asymptotes. The transverse axis is

y = x

and the conjugate axis is

y = - x

. This is exactly the "tilted head" trap.

The curve

y = 1/ x

is a rectangular hyperbola.

True. It is

x y = 1

, i.e.

x y = c^{2}

with

c = 1

. So the humble reciprocal graph you already know is a rectangular hyperbola — see Reciprocal function y=1/x.

The graph

x y = c^{2}

lives only in quadrants I and III.

True (with $c > 0$ ). Since

c^{2} > 0

, the product

x y

must be positive, so

x, y

share a sign — that's exactly quadrants I and III. Quadrants II and IV would need

x y < 0

If we allowed

x y = - c^{2}

instead, the curve would swap into quadrants II and IV.

True. Now

x y < 0

, forcing opposite signs, so the branches sit in II and IV. It's the same shape, reflected across an axis — its vertices land on

y = - x

The tangent slope on

x y = c^{2}

can be positive somewhere.

False.

\frac{d y}{d x} = - y / x

, and on each branch

x, y

have the same sign (both branches obey

c > 0

), so

- y / x < 0

everywhere. The tangent slope is always negative; only the normal (the perpendicular line) has positive slope.

Both branches of

x y = c^{2}

have the same asymptotes.

True. Both the quadrant-I branch and the quadrant-III branch flatten toward

x = 0

and

y = 0

. The two axes are shared asymptotes for the whole curve; see Asymptotes of a hyperbola.

Scaling

c

up moves the vertices

(c, c)

and

(- c, - c)

farther from the centre.

True. Vertices sit at

(\pm c, \pm c)

y = x

, at distance

c 2

from the centre (the origin). Larger

c

pushes both branches outward — the curve grows but keeps its shape.

Setting

t_{1} = t_{2}

in the chord equation

x + t_{1} t_{2} y = c (t_{1} + t_{2})

gives nonsense.

False. It gives the tangent

x + t^{2} y = 2 c t

— a chord whose two endpoints have merged into one point is exactly a tangent. This is a built-in self-check (see the formula box above).

Spot the error

x y = c^{2}

has a vertical transverse axis because

y

blows up as

x \to 0

Wrong. Blowing up near

x = 0

describes the asymptote, not the transverse axis. The transverse axis is the line

y = x

through the vertices

(\pm c, \pm c)

; the vertical line is an asymptote.

"Slope of the tangent at

t

+ 1/ t^{2}

from

\frac{d y}{d x} = y / x

Wrong sign twice. Implicit differentiation of

x y = c^{2}

gives

y + x \frac{d y}{d x} = 0

, so

\frac{d y}{d x} = - y / x = - 1/ t^{2}

(as derived in the formula box). The minus was dropped. The

+ t^{2}

belongs to the normal (negative reciprocal).

"Use the ellipse-style tangent

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

for

x y = c^{2}

Wrong formula. There

(x_{1}, y_{1})

is the point of tangency and

a, b

are the ellipse semi-axes — none of which appear in

x y = c^{2}

. This curve has its own tangent

\frac{x}{t} + t y = 2 c

(equivalently

x y_{1} + y x_{1} = 2 c^{2}

in point form). No

a^{2}, b^{2}

enter.

c^{2} = a^{2}

when we rotate

x^{2} - y^{2} = a^{2}

into

x y = c^{2}

Wrong constant. The substitution gives

- 2 X Y = a^{2}

(with

X, Y

the turned coordinates), so

X Y = - a^{2} /2

and after relabelling

c^{2} = a^{2} /2

, not

a^{2}

. The factor of 2 comes from

cos 4 5^{\circ} sin 4 5^{\circ} = \frac{1}{2}

, exactly as shown in the rotation box.

"Rotating the axes changes the size of the curve, so distances are different afterward."

Wrong. Rotation is a rigid motion — it preserves every distance and the whole shape. We only relabel the same points in turned axes; the curve itself never moves.

"The parametric point

(c t, c / t)

works for all real

t

including

t = 0

Wrong. At

t = 0

we'd need

c / t

, which is undefined — the point would sit on an asymptote, and the curve never reaches there. So

t \neq = 0

always, and

t = 0

corresponds to no real point.

"On

x y = c^{2}

, the point

(0, 0)

is the vertex since it's the centre."

Wrong. The origin is the centre (the crossing point of the asymptotes), but the curve doesn't pass through it —

0 \cdot 0 = 0 \neq = c^{2}

. The vertices are

(c, c)

and

(- c, - c)

, the nearest curve points to that centre.

Why questions

Why must

x

and

y

always share the same sign on

x y = c^{2}

Because

c^{2} > 0

and their product equals it; two numbers whose product is positive must be both positive or both negative — never one of each.

Why is the rectangle from any curve point to the axes always the same area?

Its sides are

∣ x ∣

and

∣ y ∣

, and their product

∣ x y ∣ = c^{2}

is fixed by the equation. That constant-area rule literally is the curve — it's why the equation reads

x y = c^{2}

(see s01).

Why does the naive slope formula never give a horizontal tangent for

x y = c^{2}

A horizontal tangent needs

\frac{d y}{d x} = - y / x = 0

, i.e.

y = 0

; but

y = 0

is an asymptote the curve never reaches. So the slope approaches

0

only as

x \to \infty

, never attaining it.

Why do we parametrize as

(c t, c / t)

rather than solving

y = c^{2} / x

directly?

One parameter

t

names each point symmetrically and turns chord/tangent problems into clean symmetric algebra (like the chord

x + t_{1} t_{2} y = c (t_{1} + t_{2})

), avoiding messy

c^{2} / x

substitutions.

Why is the eccentricity forced to be exactly

2

and never anything else?

Rectangular means

a = b

; eccentricity

e = 1 + b^{2} / a^{2} = 1 + 1 = 2

. The equal-axes condition pins it — you cannot tune it with

c

Why do we say the asymptotes are "perpendicular" for a rectangular hyperbola?

In the

x y = c^{2}

frame the asymptotes are the two coordinate axes, which meet at

9 0^{\circ}

. Perpendicular asymptotes are the defining feature that makes the hyperbola "rectangular" (equilateral).

Why does the tangent's intercept product on the axes stay tied to

c

The tangent

\frac{x}{t} + t y = 2 c

meets the axes at

(2 c t, 0)

and

(0, 2 c / t)

; the curve point

(c t, c / t)

is their exact midpoint — a signature property of

x y = c^{2}

tangents.

Edge cases

What happens to the point

(c t, c / t)

t \to \infty

x = c t \to \infty

while

y = c / t \to 0

: the point slides far out along the branch, hugging the x-axis asymptote. It never lands on

y = 0

What happens as

t \to 0^{+}

x = c t \to 0

while

y = c / t \to \infty

: the point climbs up hugging the y-axis asymptote. Again it never touches

x = 0

; the asymptote is approached, not reached.

What does negative

t

(say

t < 0

) describe?

With

t < 0

both

c t

and

c / t

are negative (since

c > 0

), placing the point in quadrant III — the other branch. So

t > 0

traces branch I,

t < 0

traces branch III.

c = 0

a valid rectangular hyperbola?

No.

x y = 0

means

x = 0

y = 0

— that's the two asymptote lines themselves (a degenerate pair), not a hyperbola. A genuine curve needs

c \neq = 0

(and we take

c > 0

At the vertex

(c, c)

, what is the tangent slope, and does it match the general formula?

The vertex is

t = 1

, so slope

= - 1/ t^{2} = - 1

. The tangent there is perpendicular to the transverse axis

y = x

(whose slope is

+ 1

) — exactly as a vertex tangent should be.

If two curve points have parameters

t

and

- t

, where do they sit relative to each other?

They are

(c t, c / t)

and

(- c t, - c / t)

— a reflection through the origin (the centre). So

t

and

- t

give a pair of diametrically opposite points, one per branch.

Recall One-line survival summary

Same-sign product ⇒ quadrants I & III (since $c > 0$ ); centre = origin (where asymptotes cross); axes are asymptotes, transverse axis is $y = x$ ; vertices $(\pm c, \pm c)$ are the nearest curve points to the centre; slope always negative $- 1/ t^{2}$ ; normal (perpendicular line) has slope $+ t^{2}$ ; $t \neq = 0$ , $c > 0$ ; $e = 2$ locked in.

Connections

Parent: Rectangular hyperbola xy = c² — the full derivations these traps guard.
Hyperbola standard form — the $a = b$ origin of the shape.
Rotation of axes — where the factor $c^{2} = a^{2} /2$ comes from.
Asymptotes of a hyperbola — why the axes double as asymptotes.
Tangent and Normal to conics — the tangent/normal slope traps.
Eccentricity — why $e = 2$ is forced.
Reciprocal function y=1/x — the simplest rectangular hyperbola.