Conic Sections

From a point outside a parabola you can draw exactly two tangents. How many normals can you draw?

Symmetry suggests two. The answer is three.

The normal to $y^2=4ax$ at the parameter $t$ — that is, at the point $\left(a{t}^2,2at\right)$ — has equation $y=-tx+2at+a{t}^3$. Requiring it to pass through a point $(h,k)$ gives

$a{t}^3+\left(2a-h\right)t+k=0,$

a cubic in $t$. A cubic has three roots, so up to three normals pass through a general point, and at least one always does.

The absent $t^2$ term is worth more than the count. It says that if three normals from a point meet the parabola at parameters $t_1,t_2,t_3$, then

$t_1+t_2+t_3=0,$

which turns most "concurrent normals" questions into one line of Vieta's relations rather than a page of coordinates.

The asymmetry between tangents and normals is the character of this chapter. Tangency is a quadratic condition and gives two of everything; normality is a cubic condition and gives three. Advanced sets questions on the difference.

1. One definition, three curves

A conic is the locus of a point whose distance from a fixed focus is $e$ times its distance from a fixed directrix. The single number $e$ decides everything: $e<1$ gives an ellipse, $e=1$ a parabola, and $e>1$ a hyperbola.

Two more quantities are read off the same standard forms and are worth tabulating alongside. The latus rectum, the focal chord perpendicular to the axis, has length $4a$ for the parabola and $\frac{2b^2}{a}$ for both the ellipse and the hyperbola. The directrices are $x=-a$ for the parabola and $x=\pm \frac{a}{e}$ for the other two, so they move outwards as the eccentricity falls and recede to infinity as a conic approaches a circle.

That last observation is the reason a circle has no directrix: it is the limiting ellipse with $e=0$, and the focus-directrix definition degenerates. Questions that quote a directrix are therefore telling you the conic is not a circle, which occasionally settles a case distinction on its own.

The two closed conics also have a focal distance property that is often faster than any equation: for an ellipse the two focal distances sum to $2a$, and for a hyperbola they differ by $2a$ in absolute value. A locus question phrased in terms of distances to two fixed points is usually asking for one of these directly.

2. The parabola through its parameter

Writing points as $\left(a{t}^2,2at\right)$ converts every question about the parabola into algebra in $t$. The chord joining $t_1$ and $t_2$ has slope $\frac{2}{t_1+t_2}$, and passes through the focus exactly when

$t_1{t}_2=-1.$

That single condition generates most focal-chord results. The length of a focal chord through parameter $t$ is $a{\left(t+\frac{1}{t}\right)}^2$, minimised at $t=\pm 1$ where it equals $4a$: the latus rectum is the shortest focal chord.

Illustration 1

Show that the semi-latus rectum of a parabola is the harmonic mean of the two segments of any focal chord.

With $t_1{t}_2=-1$, the focal distances of the two ends are $r_1=a\left(1+t_1^2\right)$ and $r_2=a\left(1+t_2^2\right)=a\left(1+\frac{1}{t_1^2}\right)$. Then

$\frac{1}{r_1}+\frac{1}{r_2}=\frac{1}{a\left(1+t_1^2\right)}+\frac{t_1^2}{a\left(1+t_1^2\right)}=\frac{1}{a},$

so the harmonic mean of $r_1$ and $r_2$ is $2a$, which is exactly the semi-latus rectum. Every $t_1$ cancels, which is why the result holds for every focal chord.

Illustration 2

A chord of $y^2=4ax$ subtends a right angle at the vertex. Show that it passes through a fixed point.

The chord joins $\left(a{t}_1^2,2a{t}_1\right)$ and $\left(a{t}_2^2,2a{t}_2\right)$, and the lines from the origin to these points have slopes $\frac{2}{t_1}$ and $\frac{2}{t_2}$. Perpendicularity gives $t_1{t}_2=-4$.

The chord itself has equation $2x-\left(t_1+t_2\right)y+2a{t}_1{t}_2=0$, so substituting $t_1{t}_2=-4$ makes it $2x-\left(t_1+t_2\right)y-8a=0$, which passes through $(4a,0)$ whatever $t_1+t_2$ may be. Recognising that a varying coefficient multiplies a quantity that can be made zero is the standard route to a fixed point.

3. Tangents and normals

For $y^2=4ax$ the tangent at $t$ is $ty=x+a{t}^2$ and the normal is $y=-tx+2at+a{t}^3$. In slope form these become

$y=mx+\frac{a}{m}\text{(tangent)},y=mx-2am-a{m}^3\text{(normal)}.$

The tangent form fails at $m=0$, and the failure is instructive: a line parallel to the axis meets the parabola at exactly one point but is not a tangent, since it crosses the curve. "Meets in one point" and "is a tangent" are different statements for a parabola, and only the second is what a tangency condition tests.

The corresponding slope forms for the other conics are $y=mx\pm \sqrt{a^2{m}^2+b^2}$ for the ellipse and $y=mx\pm \sqrt{a^2{m}^2-b^2}$ for the hyperbola. The minus sign in the last one carries a condition: a real tangent of slope $m$ exists only when $a^2{m}^2>b^2$, which is precisely the statement that a line flatter than the asymptotes can never touch the curve.

Illustration 3

Find the locus of a point from which two perpendicular tangents are drawn to $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.

A tangent of slope $m$ through $(h,k)$ satisfies $k=mh\pm \sqrt{a^2{m}^2+b^2}$, and squaring gives

$\left(h^2-a^2\right)m^2-2hkm+\left(k^2-b^2\right)=0.$

The product of the two slopes is $-1$, so $k^2-b^2=-\left(h^2-a^2\right)$, giving

$x^2+y^2=a^2+b^2.$

This is the director circle of the ellipse. The same calculation for the hyperbola gives $x^2+y^2=a^2-b^2$, which is empty when $b>a$ — there is then no point at all from which the tangents are perpendicular.

Illustration 4

If the normals at three points of $y^2=4ax$ are concurrent, show that the sum of the ordinates of those points is zero.

The parameters satisfy $t_1+t_2+t_3=0$ from the cubic in the opening. The ordinates are $2a{t}_1$, $2a{t}_2$ and $2a{t}_3$, so their sum is $2a\left(t_1+t_2+t_3\right)=0$.

The result is one substitution away once the cubic is written down, and it is the reason the cubic is worth writing down before anything else in a normals question.

Illustration 5

Find the locus of a point from which two perpendicular tangents are drawn to $y^2=4ax$.

A tangent of slope $m$ through $(h,k)$ satisfies $k=mh+\frac{a}{m}$, that is $h{m}^2-km+a=0$. The product of the two slopes is $\frac{a}{h}$, and setting it to $-1$ gives

$h=-a.$

The locus is the line $x=-a$ — the directrix. So the parabola's director circle degenerates into a straight line, which is what happens to a circle of infinite radius, and it explains why the directrix keeps appearing in questions that never mention it.

4. The ellipse and its auxiliary circle

The parametrisation $\left(a\cos \theta ,b\sin \theta \right)$ is not the polar angle of the point. It is the eccentric angle: the angle at the centre of the corresponding point on the auxiliary circle $x^2+y^2=a^2$, obtained by projecting the ellipse point vertically outwards.

That construction explains why the ellipse behaves like a vertically compressed circle. Every vertical distance is scaled by $\frac{b}{a}$, so areas scale by the same factor — which is why the area of the ellipse is $\pi ab$ — while horizontal distances and the eccentric angle are inherited unchanged from the circle.

Illustration 6

Find the eccentricity of an ellipse whose latus rectum equals half the minor axis.

The latus rectum has length $\frac{2b^2}{a}$ and the minor axis is $2b$, so $\frac{2b^2}{a}=b$ and $b=\frac{a}{2}$. Then

$e^2=1-\frac{b^2}{a^2}=1-\frac{1}{4}=\frac{3}{4}\Rightarrow e=\frac{\sqrt{3}}{2}.$

Every ellipse question of this kind is the same two steps: convert the stated geometric fact into a relation between $a$ and $b$, then feed it into the eccentricity formula.

Illustration 7

A point moves so that the sum of its distances from $(4,0)$ and $(-4,0)$ is $10$. Find its locus.

The condition is the focal-distance property with $2a=10$ and $ae=4$, so $a=5$ and $e=\frac{4}{5}$. Then $b^2=a^2\left(1-e^2\right)=25\left(1-\frac{16}{25}\right)=9$, and the locus is

$\frac{x^2}{25}+\frac{y^2}{9}=1.$

Recognising the definition avoids squaring two surds twice, which is the alternative.

5. Reflection, and conics that are not centred at the origin

Each conic reflects light in a way that follows from its focal property. A ray travelling parallel to the axis of a parabola reflects through the focus, which is why a dish antenna concentrates a distant signal at one point.

A ray from one focus of an ellipse reflects to the other focus, so a whispering gallery carries sound from one focus to the other along every path at once. A ray aimed at one focus of a hyperbola reflects away from the other.

In practice a conic is rarely handed over in standard position. Completing the square in both variables moves the origin to the centre or vertex, after which the standard results apply unchanged, since a translation alters no distance, no angle and no eccentricity.

Illustration 8

Identify the conic $y^2-4y-8x+28=0$ and find its focus and directrix.

Completing the square in $y$,

${\left(y-2\right)}^2=8x-24=8\left(x-3\right).$

This is a parabola with vertex $(3,2)$, opening rightwards, with $4a=8$ and so $a=2$. Shifting the standard results by the vertex, the focus is at $(3+2,2)=(5,2)$ and the directrix is $x=3-2=1$.

Checking with the definition: the point $(5,6)$ lies on the curve, since $36-24-40+28=0$, and its distance from the focus is $4$ while its distance from the directrix is also $4$.

6. The hyperbola and its asymptotes

The asymptotes $y=\pm \frac{b}{a}x$ are the skeleton of a hyperbola: the curve approaches them without ever meeting them, and their equation is obtained by replacing the $1$ on the right of the standard form by $0$. The conjugate hyperbola $\frac{y^2}{b^2}-\frac{x^2}{a^2}=1$ shares those same asymptotes and occupies the other pair of regions.

A hyperbola with $a=b$ is rectangular: its asymptotes are perpendicular and $e=\sqrt{2}$. Referred to those asymptotes as axes it takes the compact form $xy=c^2$, with parametrisation $\left(ct,\frac{c}{t}\right)$ and tangent $x+t^{2}y=2ct$.

Illustration 9

Show that the product of the perpendicular distances from any point of $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ to its asymptotes is constant.

The asymptotes are $bx\pm ay=0$, so the product of the distances from $(x_1,y_1)$ is

$\frac{|b{x}_1-a{y}_1|}{\sqrt{a^2+b^2}}\cdot \frac{|b{x}_1+a{y}_1|}{\sqrt{a^2+b^2}}=\frac{|b^2{x}_1^2-a^2{y}_1^2|}{a^2+b^2}.$

But the point lies on the curve, so $b^2{x}_1^2-a^2{y}_1^2=a^2{b}^2$, and the product is $\frac{a^2{b}^2}{a^2+b^2}$, independent of the point.

Illustration 10

The tangent at a point of $xy=c^2$ meets the axes at $A$ and $B$. Show that the point of contact is the midpoint of $AB$.

The tangent at $\left(ct,\frac{c}{t}\right)$ is $x+t^{2}y=2ct$, meeting the axes at $A=(2ct,0)$ and $B=\left(0,\frac{2c}{t}\right)$. The midpoint of $AB$ is $\left(ct,\frac{c}{t}\right)$, which is the point of contact.

A corollary follows at once: the triangle cut off by the tangent has area $\frac{1}{2}\cdot 2ct\cdot \frac{2c}{t}=2c^2$, the same for every tangent.

7. Chords of contact, polars and midpoints

Writing $S$ for the conic's equation moved to one side, three results share one notation. From an external point, the chord of contact of the two tangents is $T=0$. The chord whose midpoint is $(x_1,y_1)$ is $T=S_1$. And the pair of tangents themselves is $S{S}_1=T^2$.

Here $T$ is $S$ with each square replaced by the corresponding product and each linear term averaged, exactly as for the circle.

Illustration 11

Find the chord of the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$ whose midpoint is $(2,1)$.

With $T=\frac{2x}{16}+\frac{y}{9}-1$ and $S_1=\frac{4}{16}+\frac{1}{9}-1$, the equation $T=S_1$ reads

$\frac{x}{8}+\frac{y}{9}=\frac{1}{4}+\frac{1}{9},$

which simplifies to $9x+8y=26$. Checking, the midpoint $(2,1)$ satisfies $18+8=26$, as any chord's midpoint must.

Illustration 12

Find the chord of contact of the tangents drawn from $(3,4)$ to the parabola $y^2=8x$.

Here $4a=8$, so $a=2$ and $T=y{y}_1-4\left(x+x_1\right)$ with $(x_1,y_1)=(3,4)$. Setting $T=0$,

$4y-4(x+3)=0\Rightarrow y=x+3.$

Since $S_1=16-24=-8<0$, the point lies inside the parabola and there are in fact no real tangents — so the line found is a formal chord of contact with no contact points. Checking the sign of $S_1$ before answering is the habit this question rewards.

Summary

Tangency is a quadratic condition and normality a cubic one, so an external point admits two tangents but three normals to a parabola. The cubic $a{t}^3+\left(2a-h\right)t+k=0$ has no $t^2$ term, so the parameters of three concurrent normals sum to zero — the single most useful relation in parabola questions.

All three conics come from one focus-directrix definition with $e$ deciding which, and the latus rectum and directrices are read off the same standard forms. The focal-distance properties, a sum of $2a$ for the ellipse and a difference of $2a$ for the hyperbola, answer most locus questions faster than any equation.

On a parabola, a focal chord is characterised by $t_1{t}_2=-1$, and from it follow the length $a{\left(t+\frac{1}{t}\right)}^2$ and the fact that the semi-latus rectum is the harmonic mean of the two focal segments.

The eccentric angle parametrising an ellipse belongs to the auxiliary circle, not to the ellipse point, and the ellipse is the circle with every vertical distance scaled by $\frac{b}{a}$. For a hyperbola the asymptotes are found by replacing $1$ by $0$, they are shared with the conjugate hyperbola, and the tangent slope form $y=mx\pm \sqrt{a^2{m}^2-b^2}$ carries the condition $a^2{m}^2>b^2$, which says no line flatter than an asymptote can touch the curve.

The director circle of a parabola degenerates into its directrix, which is where perpendicular tangents meet. A conic not in standard position is brought there by completing the square, since a translation changes no distance, angle or eccentricity.

Chord of contact, chord with a given midpoint and the pair of tangents are $T=0$, $T=S_1$ and $S{S}_1=T^2$ respectively, for every conic. The sign of $S_1$ decides whether the point is inside or outside, and should be checked before any of the three is quoted.