Grade 10

The three conic sections

Ellipses, hyperbolas and parabolas, obtained by slicing a cone and defined by equations in the plane.

IntuitionSlicing a cone: three curves from one shape

Take a double cone and slice it with a flat plane. Tilt the plane a little and you get a closed oval, the ellipse. Tilt it further, parallel to a slant line of the cone, and the oval opens up into a parabola ( $y^2=2px$ ). Tilt it even more, so the plane cuts both nappes of the cone, and you get the two branches of a hyperbola. All three curves — and the circle as a special ellipse — come from the same simple idea: a plane meeting a cone at different angles.

Switch between the ellipse

\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1

(with foci

F_1, F_2

and constant focal sum

PF_1+PF_2=2a

), the hyperbola (with asymptotes), and the parabola

y^2=2ax

(with focus and directrix).

SchoolCanonical equations in the plane

Definition: Ellipse

An ellipse is the set of points $P$ in the plane whose distances to two fixed points $F_1,F_2$ (the foci) have a constant sum: $PF_1+PF_2=2a$ , where $2a$ is longer than the distance $F_1F_2$ .

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

Here $a>b>0$ ; the vertices are at $(\pm a,0)$ and the foci are $F_1(-c,0), F_2(c,0)$ with $c^2=a^2-b^2$ . The number $e=c/a$ is the eccentricity, and for an ellipse $e<1$ : the closer $e$ is to $0$ , the closer the ellipse is to a circle.

Definition: Hyperbola

A hyperbola is the set of points $P$ whose distances to two fixed foci $F_1,F_2$ have a constant absolute difference: $|PF_1-PF_2|=2a$ , where $2a$ is less than $F_1F_2$ .

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

Now the foci satisfy $c^2=a^2+b^2$ , the vertices are $(\pm a,0)$ , and the two branches approach the asymptotes $y=\pm\frac{b}{a}x$ as $|x|\to\infty$ . The eccentricity satisfies $e>1$ : the sharper the branches, the larger $e$ .

Definition: Parabola

A parabola is the set of points $P$ equidistant from a fixed point $F$ (the focus) and a fixed line (the directrix). In canonical position with focus $F\left(\frac{p}{2},0\right)$ and directrix $x=-\frac{p}{2}$ , its equation is $y^2=2px$ with $p>0$ .

y^2=2px

For every point on the parabola, the distance to $F$ equals the distance to the line $x=-\frac{p}{2}$ . By convention the eccentricity of a parabola is $e=1$ , exactly between the ellipse's $e<1$ and the hyperbola's $e>1$ .

The three conics side by side
Curve	Canonical equation	Focal property	Eccentricity
Ellipse	$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$	$PF_1+PF_2=2a$	$e<1$
Hyperbola	$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$	$\|PF_1-PF_2\|=2a$	$e>1$
Parabola	$y^2=2px$	Distance to focus = distance to directrix	$e=1$

UndergraduateTheorems: focal properties from the equations

Theorem: Focal sum theorem for the ellipse

If $P(x,y)$ lies on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with foci $F_1(-c,0), F_2(c,0)$ and $c^2=a^2-b^2$ , then $PF_1+PF_2=2a$ .

Why is it true?

This is really the definition of the ellipse restated: the curve was built as the points whose two focal distances add up to a constant. The theorem shows that the constant is $2a$ , the length of the major axis, matching the algebraic equation.

Proof

Let P(x,y) be any point on the ellipse and write r1 = PF1, r2 = PF2. By the distance formula, r1^2 = (x+c)^2 + y^2 and r2^2 = (x-c)^2 + y^2, so r1^2 - r2^2 = 4cx.

Factor the left side as (r1-r2)(r1+r2) = 4cx. Since a point on the ellipse always satisfies r1+r2 = s for some positive constant s (we will confirm s = 2a below), this gives r1 - r2 = 4cx/s.

Adding and subtracting the two relations r1+r2 = s and r1-r2 = 4cx/s gives r1 = s/2 + 2cx/s and r2 = s/2 - 2cx/s. Substituting r1 back into r1^2 = (x+c)^2+y^2 and simplifying using y^2 = b^2(1-x^2/a^2) from the ellipse equation, the x^2 terms cancel when s = 2a and c^2 = a^2-b^2.

So the constant sum is s = 2a: every point of the ellipse satisfies PF1+PF2 = 2a, as claimed.

Theorem: Focal difference theorem for the hyperbola

If $P(x,y)$ lies on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ with foci $F_1(-c,0), F_2(c,0)$ and $c^2=a^2+b^2$ , then $|PF_1-PF_2|=2a$ .

Why is it true?

Just as the ellipse is built from a constant focal sum, the hyperbola is built from a constant focal difference. This theorem checks that the algebraic equation with a minus sign between the squared terms produces that constant-difference behavior, and pins the constant down to $2a$ .

Proof

Let P(x,y) be a point on the right branch (x >= a), and set r1 = PF1, r2 = PF2. As before, r1^2 - r2^2 = (x+c)^2 - (x-c)^2 = 4cx, so (r1-r2)(r1+r2) = 4cx.

On the hyperbola the difference r1-r2 stays constant in sign and, by symmetry of the two branches, equal in size to some constant d. Writing r1+r2 = 4cx/d and combining with r1-r2 = d gives r1 = d/2 + 2cx/d.

Substitute this into r1^2 = (x+c)^2+y^2 and use y^2 = b^2(x^2/a^2-1) from the hyperbola's equation. Expanding and collecting the x^2 coefficients, the equation is satisfied for all x when d = 2a and c^2 = a^2+b^2.

Hence r1-r2 = 2a on the right branch, and by the mirror-symmetric argument r2-r1 = 2a on the left branch, so in all cases |PF1-PF2| = 2a.

UndergraduateReal-World Applications and Worked Examples

Conics show up wherever a physical quantity depends on distance in a controlled way. Planets and satellites move on elliptical orbits with the attracting body at one focus (Kepler's first law). Parabolic mirrors and satellite dishes reflect every incoming ray parallel to the axis into the single focus, which is why headlight reflectors, telescope mirrors and radio dishes are shaped as parabolas. Hyperbolas appear in navigation systems such as LORAN, where the difference in arrival time from two fixed transmitters places a receiver on one branch of a hyperbola, and in the trajectories of objects on open, unbound orbits (like some comets).

Example: Foci and eccentricity of an orbit-shaped ellipse

A cross-section of a satellite's elliptical orbit (in units of thousands of km) is modeled by $\frac{x^2}{25}+\frac{y^2}{9}=1$ . Find the coordinates of the two foci and the eccentricity of the orbit.

Solution

From the equation, a^2 = 25 and b^2 = 9, so a = 5 and b = 3 (thousand km).

Using c^2 = a^2 - b^2 = 25 - 9 = 16, we get c = 4, so the foci are F1(-4,0) and F2(4,0).

The eccentricity is e = c/a = 4/5 = 0.8. Because e is fairly close to 1, this orbit is noticeably elongated rather than nearly circular; the attracting body (Earth) sits at one focus, not at the center.

Example: Focus of a parabolic satellite dish

A satellite dish's cross-section (in centimeters) satisfies $y^2=12x$ . Find the focus, and explain why every signal arriving parallel to the axis is reflected there.

Solution

Matching $y^2=12x$ against $y^2=2px$ gives $2p=12$ , so $p=6$ .

The focus of a parabola in this position is at $\left(\frac{p}{2},0\right)$ , so here it is at $(3,0)$ .

The reflective property of the parabola says that a ray traveling parallel to the axis, when it bounces off the curve, always changes direction to pass through the focus (and conversely, a signal from the focus reflects into a beam of parallel rays). This is the same geometric fact used by the ellipse and hyperbola theorems above, specialized to the case where one focus has been pushed to infinity.

So every parallel radio wave hitting the dish converges on the single point $(3,0)$ , which is where the receiver is placed.

For the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$ , what is the focal distance $c$ (from the center to a focus)?

For the hyperbola $\frac{x^2}{9}-\frac{y^2}{16}=1$ , what is the distance $c$ from the center to each focus?

Which conic section has eccentricity $e=1$ ?

A parabolic satellite dish collects radio waves arriving parallel to its axis. Where should the receiver be placed so that all reflected waves meet at one point?