Rectangular hyperbola xy = c²: Worked examples

This page is a drill of cases. The parent note main topic built the formulas; here we hit every corner where a sign flips, a point degenerates, or an exam adds a twist. First we lay out the full grid of scenarios, then we work one example per grid-cell so you never meet a situation you haven't already seen.

Everything below uses only the tools already earned in the parent note. As a one-line reminder before we start:

Here $c$ is a fixed positive number (the "size" of the hyperbola). The curve has two vertices, at $(c, c)$ and at $(- c, - c)$ , both lying on the line $y = x$ . And $t$ is just a label for a point — one number that picks out where you are on the curve.

The scenario matrix

Every problem this topic throws is one of these cells. The last column names the example that covers it.

#	Case class	What is special about it	Covered by
A	$t > 0$ — Quadrant I branch	both $c t > 0$ and $c / t > 0$	Ex 1
B	$t < 0$ — Quadrant III branch	both coordinates negative	Ex 2
C	Same-sign chord ( $t_{1}, t_{2} > 0$ )	chord lying on one branch, slope sign check	Ex 3
D	Mixed-sign chord ( $t_{1} > 0, t_{2} < 0$ )	chord crossing between the two branches	Ex 4
E	Intersection of two tangents	symmetric-sum formula, sign bookkeeping	Ex 5
F	Degenerate: $t_{1} = t_{2}$ (chord → tangent)	limiting case, self-check	Ex 6
G	Limiting: $t \to 0$ and $t \to \infty$	point escaping to an asymptote	Ex 7
H	Normal meeting the curve again	cubic in $t$ , third point	Ex 8
I	Real-world word problem	Boyle's law $P V = const$	Ex 9
J	Exam twist: locus of midpoints	eliminate the parameter	Ex 10

Every example carries a figure so the geometry is always visible.

Example 1 — Cell A · Quadrant I ( $t > 0$ )

Worked example Tangent and the two axis-intercepts at

$t = 3$ on $x y = 4$ For $x y = 4$ take $t = 3$ . Find the tangent line, and where it cuts the $x$ -axis and $y$ -axis. Show the triangle it makes with the axes.

Forecast: Guess before computing — will the midpoint of that axis-cut segment land on the curve, or somewhere else?

Read off $c$ . $c^{2} = 4 \Rightarrow c = 2$ . The point is $(c t, \frac{c}{t}) = (6, \frac{2}{3})$ . Why this step? Every formula is written in $c$ and $t$ ; we must know $c$ before substituting.
Write the tangent with $\frac{x}{t} + t y = 2 c$ : $\frac{x}{3} + 3 y = 4$ . Why this step? This is the earned tangent form — no re-differentiating.
$x$ -intercept (set $y = 0$ ): $\frac{x}{3} = 4 \Rightarrow x = 12$ , giving $(12, 0)$ . $y$ -intercept (set $x = 0$ ): $3 y = 4 \Rightarrow y = \frac{4}{3}$ , giving $(0, \frac{4}{3})$ . Why this step? Intercepts are the natural "corners" of the tangent-with-axes triangle.
Midpoint of intercepts: $(\frac{12 + 0}{2}, \frac{0 + 4/3}{2}) = (6, \frac{2}{3})$ . Why this step? Compare it to the point of tangency from step 1.

Answer: Tangent $x + 9 y = 12$ ; intercepts $(12, 0)$ and $(0, \frac{4}{3})$ ; their midpoint is $(6, \frac{2}{3})$ — exactly the point of contact. (So the forecast answer is: yes, the point of contact bisects the tangent's intercept segment — a signature property.)

Verify: Point on line? $6 + 9 \cdot \frac{2}{3} = 6 + 6 = 12$ ✓. Point on curve? $6 \cdot \frac{2}{3} = 4 = c^{2}$ ✓.

Example 2 — Cell B · Quadrant III ( $t < 0$ )

Worked example Tangent at

$t = - 2$ on $x y = 4$ — watch the signs Same curve $x y = 4$ , but now $t = - 2$ . Find the point and tangent, and confirm the slope is still negative.

Forecast: The point is in Quadrant III (both coords negative). Will the tangent's slope be positive or negative there?

Point: $(c t, \frac{c}{t}) = (2 (- 2), \frac{2}{- 2}) = (- 4, - 1)$ . Why this step? With $t < 0$ both $c t$ and $c / t$ come out negative — that is precisely the third quadrant branch.
Slope: $- \frac{1}{t ^{2}} = - \frac{1}{( - 2 ) ^{2}} = - \frac{1}{4}$ . Why this step? $t^{2} > 0$ always, so the slope is negative on both branches — a common exam trap (mistake (b) in the parent note).
Tangent: $\frac{x}{t} + t y = 2 c \Rightarrow \frac{x}{- 2} + (- 2) y = 4$ , multiply by $- 2$ : $x + 4 y = - 8$ . Why this step? Keep the sign of $t$ inside the formula — do not "make it positive."

Answer: Point $(- 4, - 1)$ , slope $- \frac{1}{4}$ , tangent $x + 4 y = - 8$ .

Verify: On curve: $(- 4) (- 1) = 4 = c^{2}$ ✓. On line: $- 4 + 4 (- 1) = - 8$ ✓. Slope from line: $y = - \frac{1}{4} x - 2$ , slope $- \frac{1}{4}$ ✓.

Example 3 — Cell C · Same-sign chord (one branch)

Worked example Chord joining

$t_{1} = 1$ and $t_{2} = 2$ on $x y = 4$ (both in Q I) Find the chord line connecting two points on the same first-quadrant branch, and note its slope sign.

Forecast: Both points sit on the descending Q I branch. Positive or negative slope?

Endpoints: $t_{1} = 1 \Rightarrow (2, 2)$ ; $t_{2} = 2 \Rightarrow (4, 1)$ . Why this step? Both coordinates positive — a chord living entirely on one branch.
Chord formula $x + t_{1} t_{2} y = c (t_{1} + t_{2})$ with $t_{1} t_{2} = (1) (2) = 2$ and $t_{1} + t_{2} = 3$ : $x + 2 y = 2 (3) = 6.$ Why this step? The symmetric chord form again, now with a positive product.
Slope: $y = - \frac{1}{2} x + 3$ , slope $- \frac{1}{2}$ . Why this step? $- \frac{1}{t _{1} t _{2}} = - \frac{1}{2}$ — negative because $t_{1} t_{2} > 0$ . On a single branch the curve only ever falls, so a same-sign chord always has negative slope.

Answer: Chord $x + 2 y = 6$ , slope $- \frac{1}{2}$ .

Verify: $(2, 2)$ : $2 + 4 = 6$ ✓. $(4, 1)$ : $4 + 2 = 6$ ✓.

Example 4 — Cell D · Mixed-sign chord (crosses branches)

Worked example Chord joining

$t_{1} = 1$ (Q I) and $t_{2} = - 2$ (Q III) on $x y = 4$ Find the chord line connecting a point on the first-quadrant branch to one on the third-quadrant branch.

Forecast: A line joining the two opposite branches — do you expect a positive or negative slope? (Trust your picture, not a formula, first.)

Endpoints: $t_{1} = 1 \Rightarrow (2, 2)$ ; $t_{2} = - 2 \Rightarrow (- 4, - 1)$ . Why this step? We need the two actual points to sanity-check the chord equation.
Chord formula $x + t_{1} t_{2} y = c (t_{1} + t_{2})$ with $t_{1} t_{2} = (1) (- 2) = - 2$ and $t_{1} + t_{2} = - 1$ : $x - 2 y = 2 (- 1) = - 2.$ Why this step? The symmetric chord form saves the two-point-slope grind.
Slope from it: $y = \frac{1}{2} x + 1$ , slope $+ \frac{1}{2}$ . Why this step? Here $- \frac{1}{t _{1} t _{2}} = - \frac{1}{- 2} = + \frac{1}{2}$ — a positive slope because $t_{1} t_{2} < 0$ . Matches the picture: a line climbing from Q III up to Q I.

Answer: Chord $x - 2 y = - 2$ , slope $+ \frac{1}{2}$ .

Verify: $(2, 2)$ : $2 - 4 = - 2$ ✓. $(- 4, - 1)$ : $- 4 + 2 = - 2$ ✓.

Example 5 — Cell E · Intersection of two tangents

Worked example Where do the tangents at

$t_{1} = 1$ and $t_{2} = 4$ on $x y = 4$ meet? Find the meeting point of the two tangent lines.

Forecast: Both $t$ 's are positive, so both contact points are in Q I. Will their tangents cross inside Q I too?

Use the recalled intersection formula $(\frac{2 c t _{1} t _{2}}{t _{1} + t _{2}}, \frac{2 c}{t _{1} + t _{2}})$ (sixth formula above) with $c = 2, t_{1} t_{2} = 4, t_{1} + t_{2} = 5$ : $(\frac{2 \cdot 2 \cdot 4}{5}, \frac{2 \cdot 2}{5}) = (\frac{16}{5}, \frac{4}{5}) .$ Why this step? This is the ready-made result; deriving from scratch every time wastes exam minutes.
Cross-check by intersecting the lines directly. Tangents: $\frac{x}{1} + 1 \cdot y = 4$ , i.e. $x + y = 4$ ; and $\frac{x}{4} + 4 y = 4$ , i.e. $x + 16 y = 16$ . Subtract: $15 y = 12 \Rightarrow y = \frac{4}{5}$ , then $x = 4 - \frac{4}{5} = \frac{16}{5}$ . Why this step? Two independent routes agreeing is our proof.

Answer: They meet at $(\frac{16}{5}, \frac{4}{5})$ — inside Q I, as forecast.

Verify: $\frac{16}{5} + \frac{4}{5} = \frac{20}{5} = 4$ ✓ (on first tangent); $\frac{16}{5} + 16 \cdot \frac{4}{5} = \frac{16}{5} + \frac{64}{5} = \frac{80}{5} = 16$ ✓ (on second).

Example 6 — Cell F · Degenerate limit $t_{1} = t_{2}$ (chord becomes tangent)

Worked example Force the chord through

$t_{1} = t_{2} = 3$ and confirm it is the tangent Take the chord formula and let both parameters merge to $t = 3$ on $x y = 4$ . Show it collapses to the tangent at $t = 3$ .

Forecast: A chord needs two points. If they merge into one, geometrically the secant line should rotate into the tangent. Do the algebra agree?

Chord $x + t_{1} t_{2} y = c (t_{1} + t_{2})$ with $t_{1} = t_{2} = 3$ , $c = 2$ : $t_{1} t_{2} = 9$ , $t_{1} + t_{2} = 6$ , so $x + 9 y = 12.$ Why this step? Setting the two labels equal is exactly the limiting operation "second point slides onto the first."
Tangent at $t = 3$ (from Ex 1): $\frac{x}{3} + 3 y = 4$ , multiply by $3$ : $x + 9 y = 12$ . Why this step? If the chord limit is meaningful it MUST reproduce the tangent — and it does, identically.

Answer: Chord $\to$ Tangent gives the same line $x + 9 y = 12$ . The degenerate case is consistent.

Verify: Point of contact $(6, \frac{2}{3})$ on $x + 9 y = 12$ : $6 + 9 \cdot \frac{2}{3} = 12$ ✓.

Example 7 — Cell G · Limits $t \to 0$ and $t \to \infty$ (escaping to an asymptote)

Worked example What happens to the point

$(c t, c / t)$ at the extremes of $t$ ? ( $x y = 4$ ) Describe the point as $t \to 0^{+}$ , $t \to + \infty$ , and the negative mirrors, and connect this to the asymptotes.

Forecast: Both asymptotes are the coordinate axes $x = 0$ and $y = 0$ . Which extreme of $t$ hugs which axis?

As $t \to + \infty$ : $c t = 2 t \to + \infty$ but $\frac{c}{t} = \frac{2}{t} \to 0^{+}$ . The point races rightward along $y = 0^{+}$ — it approaches the $x$ -axis (an asymptote) from above. Why this step? This is why $y = 0$ is an asymptote: the curve gets arbitrarily close but $y$ never equals $0$ (would need $x = \infty$ ).
As $t \to 0^{+}$ : $c t = 2 t \to 0^{+}$ but $\frac{c}{t} = \frac{2}{t} \to + \infty$ . The point shoots up along $x = 0^{+}$ — approaching the $y$ -axis asymptote. Why this step? Symmetric reason: $x = 0$ is the other asymptote.
Negative mirror ( $t \to 0^{-}$ , $t \to - \infty$ ): same behaviour reflected through the origin, hugging the axes from the Q III side. Why this step? Covers the whole matrix cell — no extreme left unshown.

Answer: $t \to \pm \infty$ presses the curve onto the $x$ -axis; $t \to 0^{\pm}$ presses it onto the $y$ -axis. The two coordinate axes are the asymptotes.

Verify (numeric squeeze): at $t = 1000$ : point $(2000, 0.002)$ , product $= 4$ ✓ and $y$ tiny. At $t = 0.001$ : point $(0.002, 2000)$ , product $= 4$ ✓ and $x$ tiny.

Example 8 — Cell H · Normal meeting the curve again (cubic gives the third point)

Worked example The normal at

$t = 1$ on $x y = 4$ meets the curve again at parameter $t^{'}$ — find it A normal is drawn at $t = 1$ . It cuts the hyperbola a second time. Find that other parameter and point.

Forecast: A line through a hyperbola meets it in (at most) two points, so exactly one other $t^{'}$ should appear. Guess: bigger or smaller than $1$ ?

Normal at $t = 1$ : $x t^{3} - y t = c (t^{4} - 1)$ with $t = 1, c = 2$ gives $x (1) - y (1) = 2 (1 - 1) = 0$ , i.e. $y = x$ . Why this step? At the vertex $t = 1$ the point is $(2, 2)$ on the axis $y = x$ , so the normal is the axis itself — a clean, checkable case.
Substitute the parametric point $(2 t^{'}, 2/ t^{'})$ into $y = x$ : $\frac{2}{t ^{'}} = 2 t^{'} \Rightarrow t^{'2} = 1 \Rightarrow t^{'} = \pm 1$ . Why this step? Any second intersection must satisfy both the line and the parametrization.
Discard $t^{'} = 1$ (that is the starting point). The other root is $t^{'} = - 1$ , giving $(- 2, - 2)$ . Why this step? The starting parameter is always one root of the intersection equation; the new information is the second root.

Answer: The normal at $t = 1$ (the line $y = x$ ) meets the curve again at $t^{'} = - 1$ , point $(- 2, - 2)$ — the opposite vertex.

Verify: $(- 2, - 2)$ on curve: $(- 2) (- 2) = 4$ ✓; on $y = x$ : $- 2 = - 2$ ✓.

General fact (worth knowing): the normal at $t$ meets the curve again at $t^{'} = - \frac{1}{t ^{3}}$ ; at $t = 1$ that is $- 1$ , matching us exactly.

Example 9 — Cell I · Real-world (Boyle's Law is a rectangular hyperbola)

Worked example Gas at constant temperature:

$P V = 48$ (units: kPa·L) A fixed amount of gas at constant temperature obeys Boyle's Law: pressure $\times$ volume is constant, $P V = 48$ . (a) If $V = 6 L$ , find $P$ . (b) The engineer needs the "rate" $\frac{d P}{d V}$ at that state. (c) Interpret the constant-area property physically.

Forecast: $P V = const$ is literally $x y = c^{2}$ with $x = V, y = P$ , $c^{2} = 48$ . So $d P / d V$ should be negative (squeeze the gas, pressure rises).

Find $P$ : $P = \frac{48}{V} = \frac{48}{6} = 8 kPa$ . Why this step? This is just the reciprocal-function value $y = c^{2} / x$ (see Reciprocal function y=1/x).
Rate: with $P V = 48$ , implicit differentiation gives $\frac{d P}{d V} = - \frac{P}{V} = - \frac{8}{6} = - \frac{4}{3} kPa/L$ . Why this step? Same $\frac{d y}{d x} = - \frac{y}{x}$ as the curve — the physics inherits the geometry. The minus sign means compressing (smaller $V$ ) raises $P$ .
Constant-area meaning: the rectangle from the state point $(V, P)$ back to the axes has area $V \cdot P = 48$ — the "same-area rule" is exactly Boyle's constant. Why this step? It ties the abstract rectangle property to a measured physical invariant.

Answer: $P = 8 kPa$ , $\frac{d P}{d V} = - \frac{4}{3} kPa/L$ , and the state-rectangle area is the fixed $48$ .

Verify: $6 \times 8 = 48$ ✓ (units kPa·L consistent). Rate sign negative as expected ✓.

Example 10 — Cell J · Exam twist: locus of the midpoint

Worked example Midpoint of a chord of

$x y = 4$ that touches parameters $t_{1}, t_{2}$ with $t_{1} t_{2} = 1$ — find its locus As $t_{1}, t_{2}$ vary but always with product $t_{1} t_{2} = 1$ , the chord's midpoint traces some curve. Find its equation.

Forecast: A constraint like $t_{1} t_{2} = 1$ usually forces the midpoint onto a straight line or a scaled hyperbola. Which?

Midpoint coordinates: $x_{m} = \frac{c t _{1} + c t _{2}}{2} = \frac{c ( t _{1} + t _{2} )}{2}, y_{m} = \frac{c / t _{1} + c / t _{2}}{2} = \frac{c ( t _{1} + t _{2} )}{2 t _{1} t _{2}} .$ Why this step? Average the two parametric points; write everything in the symmetric quantities $t_{1} + t_{2}$ and $t_{1} t_{2}$ .
Apply the constraint $t_{1} t_{2} = 1$ : then $y_{m} = \frac{c ( t _{1} + t _{2} )}{2 \cdot 1} = \frac{c ( t _{1} + t _{2} )}{2} = x_{m}$ . Why this step? Substituting the given product collapses $y_{m}$ directly onto $x_{m}$ .

Answer: The midpoint satisfies $y_{m} = x_{m}$ : the locus is the line $y = x$ — the transverse axis of the hyperbola. (So the forecast "straight line" wins.)

Verify (numeric): take $t_{1} = 2, t_{2} = \frac{1}{2}$ (product $1$ ), $c = 2$ : points $(4, 1)$ and $(1, 4)$ , midpoint $(\frac{5}{2}, \frac{5}{2})$ — lies on $y = x$ ✓.

Recall Which cell was hardest — quick self-quiz

Chord joining $t_{1} > 0, t_{2} < 0$ has slope of what sign? ::: Positive, since slope $= - \frac{1}{t _{1} t _{2}}$ and $t_{1} t_{2} < 0$ . Chord with both $t_{1}, t_{2} > 0$ has slope of what sign? ::: Negative, since $t_{1} t_{2} > 0$ ; a single branch only falls. As $t \to \infty$ on $x y = c^{2}$ the point hugs which axis? ::: The $x$ -axis ( $y \to 0$ ). Normal at $t$ re-meets the curve at what parameter? ::: $t^{'} = - \frac{1}{t ^{3}}$ . Setting $t_{1} = t_{2}$ in the chord formula gives? ::: The tangent at that $t$ . Boyle's law $P V =$ const is which conic? ::: A rectangular hyperbola $x y = c^{2}$ .

Connections

Parent: Rectangular hyperbola xy = c² — all formulas drilled here.
Reciprocal function y=1/x — the Boyle's-law example is a scaled version of this graph.
Tangent and Normal to conics — Examples 1, 2, 8 exercise the tangent/normal machinery.
Asymptotes of a hyperbola — Example 7 shows the limiting escape onto them.
Hyperbola standard form · Rotation of axes · Eccentricity — background for why these formulas hold.