11.2 Ellipses: foci, reflection, and tangent geometry

Several descriptions of the same ellipse

An ellipse can be specified by a focus and a directrix, by two foci and a constant sum of distances, or by a quadratic equation. Each description makes a different question easier. The distance definitions explain which points belong to the curve and why reflection connects the foci. The equation and parametrization give efficient tangent calculations. We will prove the connections, including the sign conditions needed when distances are squared.

We use a centered ellipse whose major axis is horizontal. Throughout, $0\lt e\lt1$ and $a>b>0$. The letters $a,b$ denote semiaxis lengths, not full axis lengths. Put

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

The strict inequalities give two distinct foci inside a nondegenerate ellipse. A vertical major axis is obtained by exchanging the coordinate roles; the formulas below consistently use the horizontal convention. A translated or rotated ellipse retains its distance and reflection properties, although its coordinate equation may look different.

From focus and directrix to the centered equation

Start in coordinates $(\bar x,\bar y)$ with focus $(k,0)$, where $k>0$, and directrix $\bar x=0$. The focus/directrix distance condition is

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

Both sides are nonnegative, so squaring is equivalent to the original condition. Expansion must retain the constant term:

$$(1-e^2)\bar x^2-2k\bar x+\bar y^2+k^2=0.$$

Complete the square, using $1-e^2>0$:

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

Define $x=\bar x-k/(1-e^2)$, $y=\bar y$, and $a=ke/(1-e^2)$. Division by the positive right-hand side gives the standard equation with $b=a\sqrt{1-e^2}$.

Definition

Standard ellipse and its focal data

The centered ellipse is

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

Its foci are $F_1=(-c,0)$ and $F_2=(c,0)$, where $c=ae=\sqrt{a^2-b^2}$. The corresponding directrices are $D_1:x=-a/e$ and $D_2:x=a/e$.

Indeed, the original focus becomes $(k-k/(1-e^2),0)=(-ae,0)$, and the original directrix becomes $x=-k/(1-e^2)=-a/e$. Reflecting across the vertical axis gives the second focus/directrix pair and the same ellipse. Since $a/e>a$, both directrices lie outside the ellipse. All steps in the derivation are reversible, so the equation describes precisely the prescribed distance locus, rather than merely containing it.

The sum of distances to the two foci

Theorem

Constant-sum characterization

For $F_1=(-ae,0)$ and $F_2=(ae,0)$ with $a>0$ and $0\lt e\lt1$, a point $P=(x,y)$ lies on $E$ if and only if

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

Write $d_1=\sqrt{(x+ae)^2+y^2}$ and $d_2=\sqrt{(x-ae)^2+y^2}$. First suppose $d_1+d_2=2a$. Subtracting the squared distances gives

$$(d_1-d_2)(d_1+d_2)=4aex, \qquad d_1-d_2=2ex.$$

Division is allowed because $2a>0$. Adding and subtracting the sum and difference yield $d_1=a+ex$ and $d_2=a-ex$. Squaring the first equality and cancelling terms gives

$$(1-e^2)x^2+y^2=a^2(1-e^2),$$

which is the ellipse equation.

For the converse, suppose the ellipse equation holds. Its nonnegative square terms imply $|x|\le a$, and substitution of $y^2=b^2(1-x^2/a^2)$ gives

$$d_1^2=(a+ex)^2,\qquad d_2^2=(a-ex)^2.$$

The quantities $a\pm ex$ are at least $a-e|x|\ge a(1-e)>0$. Taking nonnegative square roots therefore gives the actual distances $d_1=a+ex$, $d_2=a-ex$, whose sum is $2a$. This positivity check is necessary: squaring alone would not identify the signs of the distances. It also shows that no point on the ellipse is a focus.

Concept lensGeometric

A distance ratio and a distance sum locate the same point

On the ellipse the distance to $D_1$ is $x+a/e$, so multiplying it by $e$ gives $a+ex=d_1$. The distance to $D_2$ is $a/e-x$, giving $a-ex=d_2$. The directrix descriptions compare one focal distance with a line distance; adding their expressions gives the constant-sum description. The ratio viewpoint supplies the eccentricity, while the sum viewpoint makes the full major-axis length $2a$ visible.

Parametrization, tangents, and normals

The parametrization

$$\gamma(\theta)=(a\cos\theta,b\sin\theta),\qquad 0\le\theta\le2\pi$$

covers the ellipse: normalized coordinates $(x/a,y/b)$ lie on the unit circle and therefore have such an angle. Differentiation gives tangent vector $(-a\sin\theta,b\cos\theta)$. It never vanishes because sine and cosine cannot both be zero and both semiaxis lengths are positive.

Theorem

Tangent and normal without slope exceptions

At $P=(x_0,y_0)\in E$, a normal vector is $\mathbf n=(x_0/a^2,y_0/b^2)$. The tangent line is

$$\frac{xx_0}{a^2}+\frac{yy_0}{b^2}=1.$$

The normal line is $(x,y)=(x_0,y_0)+s\mathbf n$ for $s\in\mathbb R$.

The normal is nonzero because the center is not on the ellipse. At the parametrized point it is $(\cos\theta/a,\sin\theta/b)$, whose dot product with the tangent vector is zero. Hence the tangent line is $\mathbf n\cdot((x,y)-(x_0,y_0))=0$. The constant on rearrangement is $x_0^2/a^2+y_0^2/b^2=1$, proving the displayed equation.

Implicit differentiation gives slope $-b^2x_0/(a^2y_0)$ only when $y_0\ne0$. The vector argument includes the missing vertical tangents. Similarly, the normal-line formula does not divide by $x_0$ or $y_0$, so it remains valid at all four axis vertices.

Worked example

Tangents and normals at the vertices

At $(a,0)$ the tangent equation reduces to $x=a$ and the normal line is the horizontal axis. At $(0,b)$ it reduces to $y=b$ and the normal line is the vertical axis. The other vertices give $x=-a$ and $y=-b$. For a general parameter the tangent is equivalently

$$\frac{x\cos\theta}{a}+\frac{y\sin\theta}{b}=1.$$

Substitution of $\theta=0$ and $\theta=\pi/2$ recovers the two examples. Thus no separate limiting slope calculation is needed at a vertex.

Reflection from one focus to the other

At a point $P$ of the ellipse define unit vectors

$$\mathbf u_i=\frac{P-F_i}{|P-F_i|},\qquad i=1,2.$$

They point from each focus toward $P$. Their denominators are positive by the distance formulas above. Differentiate the constant-sum identity along $P=\gamma(\theta)$. The derivative of a nonzero vector's length is its unit direction dotted with its derivative, giving

$$\gamma'(\theta)\cdot(\mathbf u_1+\mathbf u_2)=0.$$

Consequently $\mathbf N=\mathbf u_1+\mathbf u_2$ is normal to the ellipse. It cannot vanish: opposite unit vectors would place $P$ between the foci, where the distance sum is $2c\lt2a$. Thus it determines a normal direction at every ellipse point, including the axis vertices.

Proof: Distinguish incoming motion from the outgoing focal direction

An incoming ray from $F_1$ has unit velocity $\mathbf u_1$. The outgoing ray toward $F_2$ must have direction $-\mathbf u_2$, not $\mathbf u_2$. Reflection across the tangent keeps the tangential component and reverses the normal component. With $\widehat{\mathbf N}=\mathbf N/\|\mathbf N\|$, the reflected direction is

$$\mathbf u_1-2(\mathbf u_1\cdot\widehat{\mathbf N})\widehat{\mathbf N}.$$

Since both focal vectors are unit, $\mathbf u_1\cdot\mathbf N=1+\mathbf u_1\cdot\mathbf u_2$ and $\|\mathbf N\|^2=2(1+\mathbf u_1\cdot\mathbf u_2)>0$. The subtracted vector is therefore exactly $\mathbf N$, and the reflected direction is $\mathbf u_1-\mathbf N=-\mathbf u_2$. This proves the focus-to-focus reflection law with the correct direction of travel.

The same calculation expresses equality of incidence and reflection angles. It uses only a normal direction, so changing the sign of the unit normal does not change the result. At a major-axis vertex the ray returns along the axis toward the other focus; this is an included case, not an exception requiring division by a vanishing tangent slope.

Worked example

Foci above a tangent horizontal axis

Let the foci be $(9,20)$ and $(49,55)$, and suppose the ellipse is tangent to the $x$-axis at $X$. Reflect the first focus across that axis to $F_1'=(9,-20)$. Reflection gives a straight path from $F_1'$ through $X$ to $F_2=(49,55)$, so

$$2a=|XF_1|+|XF_2|=|F_1'F_2| =\sqrt{40^2+75^2}=85,\qquad a=\frac{85}{2}.$$

The line is $F_1'+t(40,75)$. Its height is zero at $t=20/75=4/15$, so $X=(59/3,0)$. The two distances are $85(4/15)=68/3$ and $85(11/15)=187/3$, adding to $85$.

Existence can also be checked, rather than merely solving for a necessary length. The actual focal separation is $\sqrt{40^2+35^2}=5\sqrt{113}\lt85$, so the distance-sum locus is a nondegenerate ellipse. For every point $Y$ on the $x$-axis, reflection and the triangle inequality give $|YF_1|+|YF_2|=|YF_1'|+|YF_2|\ge85$, with equality only at $X$. At $X$ the incident and outgoing directions are reflections in that axis, so the reflection calculation identifies it as the tangent there.

Checkpoint

Which direction leaves the reflection point toward the second focus?

Recall how the focal unit vectors were defined.

Solution · Answer

It is $-\mathbf u_2=(F_2-P)/|P-F_2|$. The vector $\mathbf u_2$ points from the second focus toward the reflection point, the opposite direction.

t = 60°

|PF₁| ≈ 4.118; |PF₂| ≈ 1.882; |PF₁| + |PF₂| = 6

On $x^2/9+y^2/4=1$, the foci are $F_1=(-\sqrt5,0)$ and $F_2=(\sqrt5,0)$. The path $F_1\to P\to F_2$ has constant length $6$. The solid incoming and dashed outgoing rays make equal angles with the tangent. The normal is parallel to $(x/9,y/4)$; reflecting the incoming direction across that tangent gives the direction toward $F_2$.

Tangents at points with perpendicular radius vectors

Let $P,Q$ move on $E$ with $OP\perp OQ$, where $O$ is the center. We seek the locus of the intersection $T$ of their tangents. The condition is on the radius vectors, not on the tangent lines themselves. Those are different conditions and must not be interchanged.

Theorem

Intersection locus for perpendicular radii

The complete locus is

$$\frac{x^2}{a^4}+\frac{y^2}{b^4}=\frac1{a^2}+\frac1{b^2}.$$

We prove both directions by recovering contact points from a proposed intersection. Write $T=(h,k)$, $\mathbf v=(h/a,k/b)$, and $M=\|\mathbf v\|^2$. A contact point can be written $(a u_1,b u_2)$, where $\mathbf u=(u_1,u_2)$ has length one. Its tangent passes through $T$ exactly when $\mathbf v\cdot\mathbf u=1$.

Let $J\mathbf v=(-v_2,v_1)$, the quarter-turn of $\mathbf v$. It is perpendicular to $\mathbf v$ and has the same length. When $M>1$ the two possible unit contact coordinates are

$$\mathbf u_\pm=\frac{\mathbf v\pm\sqrt{M-1}\,J\mathbf v}{M}.$$

To obtain this formula, the required projection onto $\mathbf v$ is $\mathbf v/M$. The remaining component must be perpendicular to $\mathbf v$; imposing unit length gives its two possible signs and the coefficient shown. Directly, the dot product with $\mathbf v$ is one, and the squared norm is $(M+(M-1)M)/M^2=1$. These are all possibilities because in the plane the perpendicular component has only one direction up to sign.

The actual radius vectors are $P=(a u_{+,1},b u_{+,2})$ and $Q=(a u_{-,1},b u_{-,2})$. Expansion yields

$$P\cdot Q =\frac{a^2+b^2-b^2h^2/a^2-a^2k^2/b^2}{M}.$$

For clarity, before simplification its numerator over $M^2$ is $a^2(v_1^2-(M-1)v_2^2)+b^2(v_2^2-(M-1)v_1^2)$. Using $v_1^2+v_2^2=M$ gives the stated expression. Thus $P\cdot Q=0$ is exactly the claimed locus equation after division by $a^2b^2$.

For necessity, perpendicular nonzero radii are neither equal nor opposite, so their tangent lines have a unique finite intersection. Two distinct unit contact vectors satisfying $\mathbf v\cdot\mathbf u=1$ require $M>1$: $M\lt1$ is impossible by the dot-product bound, and $M=1$ permits only $\mathbf u=\mathbf v$. Hence the preceding calculation applies.

For the converse, suppose $T$ satisfies the displayed locus. Then

$$M=a^2\frac{h^2}{a^4}+b^2\frac{k^2}{b^4} \ge b^2\left(\frac1{a^2}+\frac1{b^2}\right) =1+\frac{b^2}{a^2}>1.$$

The constructed contacts therefore exist and are distinct. Their tangents pass through $T$, and the computed dot product is zero. The distinct unit contact vectors cannot be opposite because both have dot product one with $\mathbf v$. They are therefore not parallel; scaling coordinates by $1/a$ and $1/b$ preserves this fact for the actual tangent normals. Hence the tangents intersect uniquely at $T$. This proves that every point of the locus is attained, including points on either coordinate axis.

Worked example

Check the locus at two vertex contacts

Choose $P=(a,0)$ and $Q=(0,b)$. Their radius vectors have dot product zero. Their tangents $x=a$ and $y=b$ meet at $T=(a,b)$, which satisfies the locus equation because its two terms become $1/a^2$ and $1/b^2$. The equation is an ellipse with unequal squared-coordinate coefficients when $a>b$; it is not the circle associated with a perpendicular-tangent condition. This vertex check verifies one attained point but does not replace the converse proof for the entire locus.

Quick checks

Checkpoint

Why does the converse distance-sum proof need a sign check?

Inspect the quantities obtained after taking square roots.

Solution · Answer

From $d_i^2=(a\pm ex)^2$ alone one gets absolute values. On the ellipse, $|x|\le a$ and $0\lt e\lt1$ imply $a\pm ex\ge a(1-e)>0$, so the distances really are $a\pm ex$.

Checkpoint

Does the tangent formula apply at a major-axis vertex?

Use the normal vector rather than a slope quotient.

Solution · Answer

Yes. At $(a,0)$ the normal is $(1/a,0)$ and the tangent is $x=a$. No division by the zero coordinate is needed.

Exercises

1. For $a>b>0$, express $e,c$ and both directrices in terms of $a,b$.
2. At $P=(a\cos\theta,b\sin\theta)$, compute both focal distances and verify their sum.
3. Find tangent and normal lines at $(-a,0)$ and $(0,-b)$.
4. In the two numerical foci example, recover the contact point and explain why the semimajor axis is $85/2$, not $85$.
5. Show algebraically that reflection in the tangent takes $\mathbf u_1$ to $-\mathbf u_2$ using $\mathbf N=\mathbf u_1+\mathbf u_2$.
6. For an intersection $T$ on the perpendicular-radius locus, explain why $M>1$ is essential and why it follows automatically from that equation.

Guided solutions

Solution · Model solution 1

$e=\sqrt{1-b^2/a^2}$, $c=\sqrt{a^2-b^2}$, and the directrices are $x=\pm a/e=\pm a^2/\sqrt{a^2-b^2}$.

Solution · Model solution 2

$d_1=a(1+e\cos\theta)$ and $d_2=a(1-e\cos\theta)$. Both are positive because $e\lt1$; their sum is $2a$.

Solution · Model solution 3

At $(-a,0)$ the tangent is $x=-a$ and the normal is $y=0$. At $(0,-b)$ the tangent is $y=-b$ and the normal is $x=0$.

Solution · Model solution 4

The reflected segment is $(9,-20)+t(40,75)$, meeting height zero at $t=4/15$, hence $X=(59/3,0)$. Its length $85$ is the full focal distance sum $2a$, so the semimajor axis is half that length.

Solution · Model solution 5

Since $\|\mathbf N\|^2=2\mathbf u_1\cdot\mathbf N>0$, $\mathbf u_1-2(\mathbf u_1\cdot\mathbf N)\mathbf N/\|\mathbf N\|^2 =\mathbf u_1-\mathbf N=-\mathbf u_2$.

Solution · Model solution 6

$M>1$ makes $\sqrt{M-1}$ positive and gives two distinct contact points. On the locus, $M\ge1+b^2/a^2>1$, as shown by weighting its two terms with $a^2$ and $b^2$. Thus no additional points must be removed.