11.3 Hyperbolas: foci, asymptotes, and tangent geometry

Motivation: a distance difference and two branches

A hyperbola differs from an ellipse in the relation between its focal distances. For an ellipse their sum is fixed; for a hyperbola the absolute value of their difference is fixed. The absolute value matters: the two branches have opposite signed differences. Removing it without specifying a branch discards half of the curve.

Definitions and the focus-directrix construction

Definition 11.3.1

(Hyperbola and its parameters)

A nondegenerate hyperbola has eccentricity $e\gt 1$. In standard horizontal form,

$$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1, \qquad a>0,\quad b>0.$$

Write

$$c=\sqrt{a^2+b^2}=ae,\qquad b=a\sqrt{e^2-1},\qquad c>a.$$

The centre is the origin, the vertices are $(\pm a,0)$, the foci are $F_1=(-c,0)$ and $F_2=(c,0)$, and their corresponding directrices are $x=-a/e$ and $x=a/e$.

To derive the equation, initially use coordinates $(\bar x,\bar y)$, a focus $(k,0)$ with $k\gt 0$, and the directrix $\bar x=0$. The focal condition is

$$\sqrt{(\bar x-k)^2+\bar y^2}=e|\bar x|.$$

Both sides are nonnegative, so squaring is reversible. Expanding and completing the square gives

$$\bar y^2=(e^2-1)\left(\bar x+\frac{k}{e^2-1}\right)^2 -\frac{k^2e^2}{e^2-1}.$$

Set

$$x=\bar x+\frac{k}{e^2-1},\quad y=\bar y,\quad a=\frac{ke}{e^2-1},\quad b=\frac{ke}{\sqrt{e^2-1}}.$$

The standard equation follows. The original focus becomes $(ae,0)$ and its directrix becomes $x=a/e$; reflection in the vertical axis supplies the other pair. The focus is separated from the directrix because the original distance between them is positive. None of these formulas involves the ellipse expression with a negative quantity under the square root.

The equation implies $|x|\geq a$. Equality gives the two vertices; otherwise there are positive and negative choices for the vertical coordinate. Thus “upper half” is not the same as “right branch”: each branch has an upper and a lower part joined at its vertex.

Theorem: the absolute focal difference

Theorem 11.3.2

(Equivalent focal and Cartesian descriptions)

Let $a\gt 0$, $c\gt a$, and $b^2=c^2-a^2$. For the two foci $(\pm c,0)$, a point belongs to the hyperbola if and only if

$$|PF_1-PF_2|=2a.$$

On the right branch $PF_1-PF_2=2a$; on the left branch the signed difference is $-2a$.

Proof: retain the branch signs when taking square roots

Suppose first that the absolute difference is given. Put $r_i=PF_i$ and choose $\sigma\in\{1,-1\}$ so that $r_1-r_2=2a\sigma$. Difference of squares gives

$$(r_1-r_2)(r_1+r_2)=4cx, \qquad r_1+r_2=2\sigma ex.$$

Therefore $r_1=\sigma(a+ex)$. Squaring this last equality and using the coordinate expression for the first distance yields

$$(x+c)^2+y^2=(a+ex)^2, \qquad (e^2-1)x^2-y^2=a^2(e^2-1).$$

This is the required equation. For the converse, begin with the standard equation and compute the squares of the two distances. It gives

$$r_1=|ex+a|,\qquad r_2=|ex-a|.$$

The absolute values cannot simply be dropped. On the right branch, $ex\geq ea=c\gt a$, so both expressions inside them are positive and their difference is twice the transverse parameter. On the left branch, $ex\leq-c$, so both are negative; taking absolute values reverses their signed difference. Both branches therefore satisfy the original condition. This proves equivalence, not merely a necessary equation obtained by squaring.

Concept lensGeometric

Two distance descriptions of the same curve

The same distance formulas also recover the focus-directrix condition:

$$r_1=e\left|x+\frac ae\right|,\qquad r_2=e\left|x-\frac ae\right|.$$

The absolute values give the distances to the corresponding vertical directrices. Each matched focus-directrix pair describes the entire hyperbola, not just the nearer branch. The absolute-difference description treats the foci symmetrically; the focus-directrix description instead singles out one focus and one line. Keeping the pairing correct is essential.

Counterexample mode

Ellipse parameters do not define a hyperbola

A claim that foci separated by four units can have a constant distance difference of six units is impossible: the reverse triangle inequality bounds the absolute distance difference by the separation of the foci. For example, $a=3,c=2$ cannot define the hyperbola above, although these parameters do allow an ellipse with distance sum six. Concretely, $P=(3,0)$ has distances five and one from the foci $(\pm2,0)$: their sum is six but their difference is four. A nondegenerate hyperbola therefore requires $c\gt a$. Equality is also excluded: equality in the reverse triangle inequality gives collinear rays, rather than a nondegenerate curve.

Worked example 11.3.3

(Identify a hyperbola and check both focal conditions)

Consider $x^2/9-y^2/16=1$. Then $a=3,b=4,c=5,e=5/3$, so the directrices are $x=\pm9/5$. The point $P=(5,16/3)$ lies on its right branch because

$$\frac{25}{9}-\frac{(16/3)^2}{16}=1.$$

Its distances to the left and right foci are respectively $34/3$ and $16/3$. Their difference is six. Its distance to the right directrix is $16/5$, which becomes $16/3$ after multiplication by the eccentricity. Thus both descriptions agree at this point. Reflecting the point across the vertical axis exchanges the two focal distances and changes only the sign of their difference.

Parametrization, tangent, and normal

The identity $\sec^2\theta-\tan^2\theta=1$ gives

$$\gamma(\theta)=(a\sec\theta,b\tan\theta).$$

Use $-\pi/2\lt\theta\lt\pi/2$ for the right branch and $\pi/2\lt\theta\lt3\pi/2$ for the left branch. The endpoint angles are excluded because their cosines vanish. On each interval the tangent coordinate ranges over all real values, while the sign of the secant chooses the branch. Hence each interval parametrizes its full branch.

Theorem 11.3.4

(Tangent and normal, including the vertices)

At $P=(x_0,y_0)$ on the hyperbola, the tangent is

$$\frac{x_0x}{a^2}-\frac{y_0y}{b^2}=1,$$

and the normal is

$$a^2y_0x+b^2x_0y-(a^2+b^2)x_0y_0=0.$$

Along the parametrization, differentiate $x(\theta)^2/a^2-y(\theta)^2/b^2=1$:

$$\left(\frac{2x}{a^2},-\frac{2y}{b^2}\right)\cdot(x',y')=0.$$

The derivative vector is nonzero, since $y'=b\sec^2\theta>0$ on both parameter intervals. Thus the vector $(2x_0/a^2,-2y_0/b^2)$ is perpendicular to the tangent. It is the gradient of $f(x,y)=x^2/a^2-y^2/b^2-1$, the vector of its two partial derivatives. It is nonzero because a point on the hyperbola cannot have zero horizontal coordinate. Expanding its point-normal equation gives the tangent formula. The normal line has this gradient direction, and its displayed equation passes through the contact point. No division by the vertical coordinate is needed.

At the vertices the tangent is vertical and the normal is the horizontal axis. Away from them, implicit differentiation gives tangent slope $b^2x_0/(a^2y_0)$ and normal slope $-a^2y_0/(b^2x_0)$. The parameter derivative remains nonzero at a vertex even though its horizontal component vanishes, so the vertical tangent is not a singularity of the curve.

Worked example 11.3.5

(Tangents at a nonvertex point and at a vertex)

At $(5,16/3)$ on the preceding hyperbola, substitution gives

$$5x-3y=9,\qquad 9x+15y=125$$

for the tangent and normal. Their slopes are $5/3$ and $-3/5$, whose product is negative one. At $(3,0)$, however, the corresponding equations are $x=3$ and $y=0$. A calculation that divides by the vertical coordinate misses this valid tangent. The implicit equations include both cases without a separate limiting convention.

Asymptotes and the sign of the square root

Solving for the vertical coordinate gives

$$y=\pm\frac ba\sqrt{x^2-a^2},\qquad |x|\geq a.$$

The difference from the absolute-value approximation is

$$\sqrt{x^2-a^2}-|x| =\frac{-a^2}{\sqrt{x^2-a^2}+|x|}\longrightarrow0 \quad\text{as }|x|\longrightarrow\infty.$$

Consequently the asymptotes are $y=\pm(b/a)x$. For the upper half, the positive-slope line applies as the horizontal coordinate tends to positive infinity, and the negative-slope line applies as it tends to negative infinity. The lower half reverses these choices. Replacing the square root of the square by the coordinate itself would incorrectly exchange the left-hand behaviour. The vertical difference tending to zero also implies perpendicular distance to the corresponding line tends to zero. An asymptote is approached at infinity; it is not a tangent at a finite point.

Reflection: distinguish a real focus from a virtual one

For the ellipse, differentiating a constant distance sum gave the normal $u_1+u_2$. A constant distance difference instead gives $u_1-u_2$, changing the reflected ray's direction.

Let $u_i=(P-F_i)/|P-F_i|$, unit vectors directed from the foci to the point. On either branch the signed distance difference is constant. Differentiating along a regular parametrization therefore shows that the tangent is perpendicular to $u_1-u_2$. This difference is nonzero: equality of the two unit directions would require the point to lie on an outer focal-axis ray, where the absolute distance difference is the full focal separation instead.

For an incoming ray travelling from the first focus towards the point, its incident direction is $u_1$. Reflection across the tangent gives

$$u_1-2\frac{u_1\cdot(u_1-u_2)}{|u_1-u_2|^2}(u_1-u_2)=u_2,$$

because the numerator before the factor two is $1-u_1\cdot u_2$, while the squared norm is twice that quantity. The outgoing ray therefore travels away from the second focus; its backward extension passes through that focus. This is the virtual-focus interpretation of equal inclination to the tangent. It is different from the ellipse's reflection towards the other focus.

At a vertex the two unit vectors are opposite. Their difference still defines a normal, and the reflection formula remains valid. Their sum is zero there, so using that sum as a tangent direction would fail precisely at the vertex.

Perpendicular feet: why a circle equation needs exclusions

Worked example 11.3.6

(Find the exact pedal locus of a focus)

Fix $F=(-c,0)$. Let $X=(u,v)$ be the perpendicular foot from this focus to a tangent at a finite hyperbola point. We first prove that every foot lies on $u^2+v^2=a^2$, and then determine which circle points are actually attained.

Write the tangent as $\alpha x+\beta y=1$. Its coefficients satisfy

$$\alpha=\frac{x_0}{a^2},\quad \beta=-\frac{y_0}{b^2},\quad a^2\alpha^2-b^2\beta^2=1.$$

Put $N=\alpha^2+\beta^2\gt 0$. Orthogonal projection gives

$$X=F+\lambda(\alpha,\beta),\qquad \lambda=\frac{1+c\alpha}{N}.$$

Taking the squared distance of this point from the origin yields

$$|X|^2=c^2+\frac{1-c^2\alpha^2}{N} =c^2-b^2=a^2.$$

The second equality uses the coefficient identity and $c^2=a^2+b^2$. This establishes containment in the circle, but containment alone does not prove the exact locus.

Conversely, choose a circle point $(u,v)$ and set $d=a^2+cu$. The vector $X-F=(u+c,v)$ is nonzero since the focus lies outside the circle. The unique line through the candidate foot perpendicular to that vector is

$$(u+c)x+vy=d.$$

If $d\ne0$, define the finite point

$$P=\left(\frac{a^2(u+c)}d,-\frac{b^2v}d\right).$$

Using the circle equation, direct expansion gives

$$a^2(u+c)^2-b^2v^2=(a^2+cu)^2=d^2.$$

Thus this point belongs to the hyperbola, and its tangent is exactly the candidate line. That line contains the chosen circle point and is perpendicular to the segment from the focus, proving that the chosen point is its foot. This explicit contact point supplies the required converse.

When $d=0$, the two candidates are

$$\left(-\frac{a^2}{c},\frac{ab}{c}\right),\qquad \left(-\frac{a^2}{c},-\frac{ab}{c}\right).$$

Their candidate lines pass through the origin and are the two asymptotes. No tangent at a finite hyperbola point passes through the origin, as the standard tangent equation shows. These two points are therefore excluded; they are only limits of feet as the contact point tends to infinity. The exact locus is the circle with those two points removed.

For $a=3,b=4,c=5$, the tangent $5x-3y=9$ has perpendicular foot $(0,-3)$ from $(-5,0)$: its normal direction is $(5,-3)$, and adding that vector to the focus reaches the foot. The excluded points are $(-9/5,\pm12/5)$, not points of contact on the hyperbola.

Common mistakes and summary

Confusing the sum with the absolute difference changes the conic. Omitting a branch sign invalidates the converse. Dividing by the vertical coordinate loses vertex tangents. Finally, an equation obtained by elimination may describe the locus together with its limit points—its closure—rather than the exact attained set.

The hyperbola's standard equation connects all these issues: it determines the branch distances, supplies regular tangent and normal equations, and proves that an asymptote cannot also be a finite-point tangent. In a locus problem, state the parameter domain and prove attainability after deriving the equation.

Quick checks

Quick check

(Why must the focal condition use an absolute difference if both branches are intended?)

Compare the focal distances on the left and right branches.

Solution · Answer

The right branch has signed difference $2a$; the left branch has signed difference $-2a$. Both have absolute difference $2a$.

Quick check

(Does a zero vertical coordinate prevent a tangent from existing?)

Use the implicit tangent equation at a vertex.

Solution · Answer

No. At $(\pm a,0)$ the tangent is $x=\pm a$, and the normal is $y=0$. Only the finite-slope formula fails.

Exercises

1. For $x^2/16-y^2/9=1$, find the foci, eccentricity, directrices, and asymptotes.
2. Find the tangent and normal at $(5,9/4)$ on that hyperbola.
3. Determine the signed focal difference on its left branch, labelling the left focus first.
4. For $a=3,b=4$, determine the exact perpendicular-foot locus from the left focus and explain why two circle points are missing.
5. Explain why a hyperbola's reflected ray in the construction above travels away from the second focus, including at a vertex.

Solution · Model solution 1

Here $a=4,b=3,c=5,e=5/4$. The foci are $(\pm5,0)$, the directrices are $x=\pm16/5$, and the asymptotes are $y=\pm3x/4$.

Solution · Model solution 2

Substitution in the tangent gives $5x-4y=16$. The normal is $16x+20y=125$. Both contain the point; their slopes are negative reciprocals.

Solution · Model solution 3

It is $-8$, since the point is closer to the left focus. Its absolute value is eight.

Solution · Model solution 4

The locus is $u^2+v^2=9$ excluding $(-9/5,\pm12/5)$. At these points the unique perpendicular candidate lines are asymptotes, not tangents at finite contacts. Every other circle point has the explicit finite contact given in the worked example.

Solution · Model solution 5

Reflection sends $u_1$ to $u_2$, the unit direction from the second focus to the contact point. The outgoing ray points away from that focus. At a vertex the two unit vectors are opposite, and the same reflection formula still holds.