Hyperbola

Take two fixed points, the foci, and ask for every point whose difference of distances from them stays the same constant number. The ellipse asked for a constant sum and closed into a loop; the hyperbola asks for a constant difference and springs open into two separate, mirror-image branches that never meet. It is the last of the standard conics on the ISC Class 11 syllabus, and the algebra needed to handle it is almost identical to the ellipse's — squaring twice to clear two square roots — except one crucial sign flips along the way. Get that sign right and the whole family of results (vertices, foci, eccentricity, latus rectum) falls out the same way it did for the ellipse, just with the roles of addition and subtraction swapped.

The core insight

A hyperbola is the locus where the difference of distances from two foci is constant, giving $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ with $c^2=a^2+b^2$ and eccentricity $e=c/a>1$ .

The core idea

Place the centre at the origin with foci on the $x$ -axis at $(\pm c,0)$ . A point $P(x,y)$ lies on the hyperbola exactly when the difference of its distances from the two foci is a fixed constant $2a$ , with $c>a>0$ :

|PF_1-PF_2|=2a

where: $F_1,F_2$ are the two foci, $PF_1$ and $PF_2$ are the distances from $P$ to each focus, and $2a$ is the constant difference (the length of the transverse axis). Clearing the two square roots by squaring twice — the same mechanical process used for the ellipse — turns this distance condition into the standard equation

\frac{x^2}{a^2}-\frac{y^2}{b^2}=1, \qquad \text{where } b^2=c^2-a^2

where: $a$ is the semi-transverse axis (distance from centre to vertex), $b$ is the semi-conjugate axis, and $c$ is the distance from centre to focus. Notice the sign is the whole story here: the ellipse's relation is $b^2=a^2-c^2$ because its foci sit inside the curve, but a hyperbola's foci sit outside the vertices, so the roles reverse and $c^2=a^2+b^2$ — addition, not subtraction. This single sign difference is what separates the closed ellipse from the open, two-branched hyperbola. Reading the standard form $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ (transverse axis along $x$ ) gives every feature at a glance: centre $(0,0)$ ; vertices $(\pm a,0)$ , the points where the curve actually touches the transverse axis; foci $(\pm c,0)$ with $c=\sqrt{a^2+b^2}$ ; eccentricity $e=\dfrac{c}{a}=\sqrt{1+\dfrac{b^2}{a^2}}$ , always greater than $1$ for a hyperbola (compare the ellipse's $e<1$ and the parabola's $e=1$ ); directrices $x=\pm\dfrac{a}{e}$ ; transverse axis of length $2a$ along the $x$ -axis and conjugate axis of length $2b$ along the $y$ -axis (the conjugate axis does not meet the curve at all — a genuinely new idea, since the ellipse's minor axis does touch it); and latus rectum $\dfrac{2b^2}{a}$ , the same formula shape as the ellipse's. Swap the equation to $\dfrac{y^2}{a^2}-\dfrac{x^2}{b^2}=1$ and the transverse axis moves to the $y$ -axis, with vertices $(0,\pm a)$ and foci $(0,\pm c)$ . The rule for spotting orientation is genuinely different from the ellipse: for an ellipse the larger denominator marks the major axis, but for a hyperbola it is purely the sign of the term — whichever variable carries the plus sign owns the transverse axis, regardless of how the two denominators compare in size. Two disconnected branches, not one closed loop, are the hyperbola's visual signature, and this falls straight out of the defining condition: points satisfying $PF_1-PF_2=2a$ cluster near one focus, forming one branch, while points satisfying $PF_2-PF_1=2a$ cluster near the other, forming the mirror branch. Between the two branches lies a gap of width $2a$ that the curve never crosses.

Figure

Two mirror-image hyperbola branches opening left and right along the x axis with the two foci marked outside the vertices and the conjugate axis shown as a dashed vertical segment that does not touch the curve

Derivation AStandard equation of a hyperbola from the focal-distance condition

The goal is to convert the geometric definition — the difference of distances from two fixed foci stays constant at $2a$ — into the algebraic standard form $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ , using only the distance formula and two rounds of squaring to eliminate the square roots, exactly mirroring the ellipse derivation but tracking a sign flip at the finish.

Step 1. Set up coordinates. Put the centre at the origin with foci $F_1(-c,0)$ and $F_2(c,0)$ , and let $P(x,y)$ be a general point on the curve. The defining condition is $PF_1-PF_2=\pm 2a$ (the sign depends on which branch $P$ sits on), so write $PF_1=\sqrt{(x+c)^2+y^2}$ and $PF_2=\sqrt{(x-c)^2+y^2}$ , and take $PF_1-PF_2=2a$ for the right-hand branch without loss of generality.

Step 2. Isolate one radical and square. Rearranging gives $\sqrt{(x+c)^2+y^2}=2a+\sqrt{(x-c)^2+y^2}$ . Squaring both sides and expanding $(x+c)^2$ and $(x-c)^2$ , most terms cancel, leaving $2cx-4a^2=4a\sqrt{(x-c)^2+y^2}$ , which simplifies to $cx-a^2=a\sqrt{(x-c)^2+y^2}$ .

Step 3. Square again to clear the last radical. $(cx-a^2)^2=a^2\left[(x-c)^2+y^2\right]$ expands to $c^2x^2-2a^2cx+a^4=a^2x^2-2a^2cx+a^2c^2+a^2y^2$ . The $-2a^2cx$ terms cancel on both sides, leaving $c^2x^2-a^2x^2-a^2y^2=a^2c^2-a^4$ , i.e. $(c^2-a^2)x^2-a^2y^2=a^2(c^2-a^2)$ .

Step 4. Name $b^2=c^2-a^2$ and divide through. Substituting gives $b^2x^2-a^2y^2=a^2b^2$ ; dividing every term by $a^2b^2$ produces $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ . The key sign check: since $c>a$ for a hyperbola, $c^2-a^2$ is positive, so $b^2$ is a genuine positive quantity — the opposite of the ellipse, where $a>c$ forces $b^2=a^2-c^2$ instead. The derivation confirms $c^2=a^2+b^2$ directly from $b^2=c^2-a^2$ , matching the relation used throughout this article.

Worked example

Find the equation of a hyperbola with vertices $(\pm 4,0)$ and foci $(\pm 5,0)$ . Find its eccentricity, latus rectum, and the length of the gap between the branches, then extend to check whether the point $(6,\sqrt{5})$ lies on the curve.

Step 1. Both vertices and foci lie on the $x$ -axis, so the transverse axis is horizontal and the form is $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ . From the vertices, $a=4$ , so $a^2=16$ ; from the foci, $c=5$ , so $c^2=25$ .

Step 2. Find $b^2$ using $b^2=c^2-a^2=25-16=9$ . The equation is

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

Step 3. Eccentricity and latus rectum: $e=\dfrac{c}{a}=\dfrac{5}{4}$ , and the latus rectum length is $\dfrac{2b^2}{a}=\dfrac{2(9)}{4}=4.5$ . The gap between the branches — the distance between the two vertices — is $2a=8$ .

Step 4 (stretch check). Test whether $(6,\sqrt5)$ satisfies the equation: $\dfrac{6^2}{16}-\dfrac{(\sqrt5)^2}{9}=\dfrac{36}{16}-\dfrac{5}{9}=2.25-0.556=1.694\neq 1$ . So the point does not lie on this hyperbola — a reminder that only points exactly satisfying the equation belong to the curve, not merely points "in the same region."

Final answer: equation $\dfrac{x^2}{16}-\dfrac{y^2}{9}=1$ , eccentricity $\dfrac{5}{4}$ , latus rectum $4.5$ , and $(6,\sqrt5)$ is not on the curve.

Common traps

Using the ellipse's relation by habit. A hyperbola obeys $c^2=a^2+b^2$ because the foci lie beyond the vertices; confusing it with the ellipse's $b^2=a^2-c^2$ is the single most common slip.
Judging orientation from denominator size. For a hyperbola the transverse axis is fixed by the sign of the term (whichever variable is positive), not by which denominator is numerically larger — the opposite rule from the ellipse.
Skipping the reduction to " $=1$ " form. An equation like $16y^2-9x^2=144$ must be divided through by the constant before $a^2$ and $b^2$ can be read off; reading coefficients straight off the unreduced form gives wrong values.
Forgetting the eccentricity range. Every hyperbola has $e>1$ ; a computed $e\le 1$ signals an arithmetic slip somewhere in $a$ , $b$ , or $c$ .

Ask the tutor: If the sign flip between the ellipse's and hyperbola's $a,b,c$ relations is still tripping you up, ask the AI tutor to walk both derivations side by side.

Why does the hyperbola have two separate branches while the ellipse is one closed loop?

It comes straight from the defining condition. The ellipse fixes the sum of distances to the foci, which keeps a point boxed into a single bounded region. The hyperbola fixes the difference, and a difference can be achieved in two distinct ways — being closer to the left focus by $2a$ , or closer to the right focus by $2a$ — producing two separate families of points, one hugging each focus. There is no way to travel continuously from one family to the other without breaking the fixed-difference condition, so the two branches never connect.

How do I quickly tell if 9x squared minus 4y squared equals 36 opens along x or y?

Divide through by the constant on the right first, here $36$ , to get $\dfrac{x^2}{4}-\dfrac{y^2}{9}=1$ . The transverse axis belongs to whichever variable is positive — here it is $x^2$ , so the hyperbola opens left-right along the $x$ -axis, with $a^2=4$ and $b^2=9$ . Never judge from the raw numerical coefficients before dividing, and never judge from which denominator ends up larger afterward — only the sign decides.

Is the conjugate axis just the hyperbola's version of the ellipse's minor axis?

Only partly. Both are the axis perpendicular to the main axis with half-length $b$ , and both feed into the same formula for the latus rectum. But the ellipse's minor axis touches the curve at two points, while the hyperbola's conjugate axis touches the curve nowhere — it is purely a construction segment used to define $b$ and to draw the asymptote rectangle. Don't expect to find the curve crossing the $y$ -axis of a horizontally opening hyperbola; it never does.

Hyperbola

The core idea

Worked example

Common traps

Before this

Where this leads