Hyperbola — standard forms, asymptotes, foci, eccentricity: Question bank

Notation reminder, so no symbol is unearned: $a$ = transverse semi-axis (centre to vertex), $b$ = conjugate semi-axis, $c$ = centre-to-focus distance, $e = c / a$ = eccentricity, foci $F_{1}, F_{2}$ , and a general point $P$ on the curve. The defining property is $P F_{1} - P F_{2} = 2 a$ .

Which axis is transverse? It is not fixed by whether $a > b$ . It is fixed by which term is positive: in $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ the positive term is $x^{2}$ , so the transverse axis is horizontal (vertices on the $x$ -axis); in $\frac{y ^{2}}{a ^{2}} - \frac{x ^{2}}{b ^{2}} = 1$ the positive term is $y^{2}$ , so the transverse axis is vertical.

True or false — justify

For an ellipse

c^{2} = a^{2} - b^{2}

, but for a hyperbola

c^{2} = a^{2} + b^{2}

True. Ellipse adds focal distances so foci sit inside (

c < a

); hyperbola subtracts so foci sit outside the vertices (

c > a

), forcing the plus sign.

A hyperbola can have eccentricity equal to

1

False.

e = c / a

with

c > a

always, so

e > 1

strictly;

e = 1

is a parabola, the boundary case a hyperbola never reaches.

The conjugate axis of length

2 b

actually touches the curve

False. The transverse axis (length

2 a

) hits the vertices; the conjugate axis lies along the direction the curve never reaches, so it only sets the box height, never meets the branches.

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

we always have

a > b

False. Unlike the ellipse, there is no size rule between

a

and

b

;

b

can be larger, equal (rectangular), or smaller. The transverse direction is fixed by which term is positive, not by which letter is bigger.

The asymptotes pass through the centre of the hyperbola

True. Setting the right side to

0

gives

\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 0

, two lines through the origin; the centre is exactly where they cross.

The curve eventually touches its asymptote if you go far enough

False. The vertical gap between the branch and the line

y = \frac{b}{a} x

shrinks toward

0

but stays strictly positive for every finite

x

— it approaches forever, like a car racing a wall it never scrapes.

Both foci lie on the transverse axis

True. The foci are the two thumbtacks whose distance-difference defines the curve; the transverse axis is precisely the axis through the vertices and both foci.

A rectangular hyperbola always has eccentricity

2

True. Rectangular means

a = b

, so

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

, and the asymptotes

y = \pm x

are perpendicular.

Increasing

b

while holding

a

fixed makes the branches narrower

False. Slope

b / a

and

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

both rise, so the branches fan out and open wider, not narrower.

Two hyperbolas with the same asymptotes must be identical

False. Many hyperbolas share asymptotes (they just differ by the "

1

" on the right, or use the conjugate form) — same guides, different sizes and even different orientation.

Spot the error

c^{2} = a^{2} - b^{2}

, so with

a = 4, b = 3

the foci are at

(\pm 7, 0)

Error: used the ellipse relation. Hyperbola uses plus:

c^{2} = 16 + 9 = 25

, foci at

(\pm 5, 0)

— farther out than the vertices, as required.

"Asymptote slope is

a / b

because

a

is written first"

Error: slope is

b / a

= rise/run. Check the box corner

(a, b)

: rise

b

over run

a

gives

b / a

"For

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

the asymptote slope is still

b / a

Error: in the vertical form the box is

y = \pm a

x = \pm b

, so corner

(b, a)

gives slope

a / b

. The slope is always the conjugate semi-axis over the transverse semi-axis, and this ratio flips when the transverse axis is vertical.

e = 1.25 < 2

, so this can't be a hyperbola"

Error: any

e > 1

is a hyperbola.

1.25 > 1

qualifies; there is no upper limit, and

2

is not a threshold.

"The transverse axis is the longer axis, like the major axis of an ellipse"

Error: 'transverse' is defined by containing the vertices and foci, not by length. It can be shorter than the conjugate axis when

a < b

"Since the '

1

' is negligible far out, the curve becomes the line and they merge"

Error: negligible is not zero. The curve approaches

\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 0

but the residual keeps a strictly positive (shrinking) gap forever.

"Foci at

(0, \pm 10)

2 a = 16

: the equation is

\frac{x ^{2}}{64} - \frac{y ^{2}}{36} = 1

Error: foci on the

y

-axis mean the vertical form, so the

y^{2}

term is positive:

\frac{y ^{2}}{64} - \frac{x ^{2}}{36} = 1

. The positive term must sit over the transverse (

y

) direction.

Why questions

Why does a hyperbola have two branches while an ellipse has one loop

The ellipse adds distances so the point stays trapped near both foci; the hyperbola takes the absolute difference, so a point can be closer to either focus — one choice per branch, giving two open arcs.

Why is the plus sign (

c^{2} = a^{2} + b^{2}

) tied to the foci being outside the vertices

Because

c > a

for a hyperbola,

c^{2} - a^{2} = b^{2} > 0

, which only makes sense as a plus relation; geometrically the focus at

c

sits beyond the vertex at

a

Why do we introduce

b

at all if only

a

and

c

appear in the locus definition

b^{2} = c^{2} - a^{2}

cleans up the equation into

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

and, more usefully,

b

turns out to be the conjugate semi-length that sets the asymptote slopes

b / a

Why does eccentricity control asymptote steepness

Because

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

and slope is

b / a

; both grow together, so a bigger

e

literally means steeper asymptotes and wider-opening branches.

Why can we drop the "

1

" to find asymptotes but not to find vertices

Far from the centre

x^{2} / a^{2}

dwarfs

1

, so ignoring it is a good approximation there; near the vertices

x^{2} / a^{2} \approx 1

, so the "

1

" is the whole story and cannot be dropped.

Why does the branch–asymptote gap actually shrink to zero

Solving gives the branch

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

, so the gap below the line is

\frac{b}{a} x (1 - 1 - a^{2} / x^{2})

; the bracket behaves like

\frac{a ^{2}}{2 x ^{2}}

for large

x

, so the gap behaves like

\frac{ab}{2 x} \to 0

— it fades like

1/ x

, never reaching

0

Why does the rectangular hyperbola have perpendicular asymptotes

With

a = b

the slopes are

+ 1

and

- 1

, whose product is

- 1

— the condition for two lines to be perpendicular.

Why is the directrix definition

P F / (dist to directrix) = e

called unifying

The same ratio

= e

describes all conics:

e < 1

ellipse,

e = 1

parabola,

e > 1

hyperbola. One rule, three shapes, distinguished only by where

e

sits relative to

1

Edge cases

What shape do the branches approach as

e \to 1^{+}

They narrow toward a single parabola-like arc;

b / a \to 0

so the asymptotes flatten toward the transverse axis and the two branches barely spread.

What happens to the branches as

e \to \infty

(with

a

fixed)

b ≫ a

, so the asymptote slope

b / a \to \infty

: the branches straighten toward two near-vertical lines and the curve opens almost completely.

\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 0

a hyperbola

No — with

0

on the right it degenerates into the pair of asymptote lines through the centre, the limiting "skeleton" the true hyperbola only approaches.

Can a hyperbola have its centre off the origin

Yes; replace

x \to x - h

y \to y - k

to shift the centre to

(h, k)

. All relations (

c^{2} = a^{2} + b^{2}

e = c / a

, slopes

b / a

) stay the same — only the location moves.

What is the smallest possible distance-difference

2 a

for given foci

A genuine hyperbola needs

0 < 2 a < 2 c

: strictly positive (else there is no fixed difference to trace) and strictly less than the focal separation. As

2 a \to 0^{+}

the two branches flatten onto the perpendicular bisector of the foci (a straight line), the degenerate boundary; as

2 a \to 2 c^{-}

the branches collapse onto the two outward rays along the focal axis.

Recall One-line self-test

Cover every answer above and re-derive the reason from the central box (rectangle with corners $(\pm a, \pm b)$ , diagonals = asymptotes) and the sign in the equation. If any item needs more than a sentence of thought, revisit that section of the parent note.