11.5 Conic loci and parametric applications

Motivation

A locus problem starts with a geometric construction and asks which points it can produce. Substitution and elimination usually supply an equation, but that equation may contain extra branches or points that the construction cannot reach. A complete solution has two directions: every constructed point satisfies the claimed conditions, and every point satisfying those conditions can be constructed.

Concept lensStructural

A parameter records more than a location

For a chord, two parameters identify its endpoints. For a tangent, one parameter identifies its contact point, including contacts where a slope is infinite. For a rolling circle, the parameter also records how far the motion has progressed. Eliminating that parameter can preserve the set of positions while losing order, direction, or admissibility. Decide which information the question requires before choosing what to eliminate.

Positive hyperbola chords and their midpoints

Consider only the positive branch $xy=1$, with $x,y>0$. Choose two distinct chord endpoints on this branch, $S=(s,1/s)$ and $T=(t,1/t)$, where $s,t>0$ and $s\ne t$. Their chord has slope $-1/(st)$, so its equation is

$$x+sty=s+t.$$

This can also be checked by substituting both endpoints. If the chord passes through $(2,2)$, then

$$2+2st=s+t,\qquad t=\frac{s-2}{2s-1}.$$

The denominator cannot vanish: substituting $s=1/2$ in the preceding relation would give a contradiction. Positivity of both parameters gives exactly

$$0\lt s\lt\frac12\quad\text{or}\quad s>2.$$

The endpoints $s=0,1/2,2$ are excluded, for different reasons: an undefined hyperbola point, an impossible chord equation, and a zero second parameter. Moreover $s=t$ would imply $s^2-s+1=0$, which has no real roots. Thus the admissible intervals automatically give two distinct endpoints.

Theorem 11.5.1

(Exact midpoint locus on the positive branch)

The midpoints of these chords through $(2,2)$ form exactly

$$(x-1)(y-1)=1,\qquad x>1,\quad y>1.$$

Worked example 11.5.2

(Eliminate the parameters and reconstruct the endpoints)

Let $M=(x,y)$ be the midpoint. Its coordinates and the chord relation give

$$x=\frac{s+t}{2}=1+st,\qquad y=\frac{s+t}{2st}=\frac{1+st}{st}.$$

Since $st>0$, both coordinates exceed one. Eliminating the product yields $(x-1)(y-1)=1$, proving necessity including the branch restriction.

Conversely, choose a point of that restricted curve. Recover $s,t$ as the roots of

$$z^2-2xz+(x-1)=0.$$

Its discriminant is $4(x^2-x+1)>0$, because $x^2-x+1=(x-1/2)^2+3/4$. The sum of its roots is positive and their product is positive, so both roots are positive and distinct. Their sum and product also give $s+t=2+2st$, proving that the reconstructed chord passes through $(2,2)$. Its midpoint has horizontal coordinate $x$ and vertical coordinate $x/(x-1)=y$, as required. This proves every point of the specified branch is attained.

For example, $s=3$ gives $t=1/5$ and midpoint $(8/5,8/3)$. The chord is $x+(3/5)y=16/5$. Its passage through $(2,2)$ and the midpoint relation are both directly checkable. The other branch of the eliminated equation does not belong to the locus because it would require a nonpositive endpoint product.

Parabola tangents without missing the vertical one

For $y^2=4x$, use contact parameter $t$ and contact point $T(t)=(t^2,2t)$. The tangent at that point is

$$x-ty+t^2=0.$$

It includes $t=0$, the vertical tangent $x=0$. A tangent through a fixed point $Q=(p,q)$ therefore corresponds exactly to a real root of

$$t^2-qt+p=0.$$

Write $\Delta=q^2-4p$. There are two distinct real tangents precisely when $\Delta>0$, one when $\Delta=0$, and none when $\Delta\lt0$. The two parameters satisfy $t_1+t_2=q$ and $t_1t_2=p$. Distinct parameters produce different tangent lines, so this root count is also a line count.

For a nonvertical tangent, $t\ne0$ and its slope is $m=1/t\ne0$. Rearrangement gives $y=mx+1/m$ and $q=pm+1/m$. One may also derive the intercept by substituting $y=mx+c$ into the parabola and setting the resulting quadratic discriminant to zero, provided $m\ne0$.

Worked example 11.5.3

(A reciprocal-slope condition with two possible intersections)

When both tangent slopes are finite, the parameter relations give

$$\left(\frac1{m_1}-\frac1{m_2}\right)^2 =(t_1-t_2)^2=(t_1+t_2)^2-4t_1t_2=q^2-4p.$$

If this value is five and $Q$ lies on $y=x$, then $p=q$ and $p^2-4p-5=0$. Thus $Q=(-1,-1)$ or $Q=(5,5)$. Both have positive discriminant five, so each really admits two distinct contacts. Their parameter products are nonzero, so neither pair includes a vertical tangent and the original reciprocal-slope expression is defined.

The same squared parameter difference is meaningful even when a contact parameter is zero. In that case, however, one tangent slope is infinite; state the result in parameter form rather than treating infinity as a real number in a reciprocal calculation.

The angle between the two contact rays

Now let $T_1,T_2$ be the contacts from an external point $Q=(p,q)$, and let $\alpha=\angle T_1QT_2$ be the angle between the rays from $Q$ to the contacts. Thus $0\lt\alpha\lt\pi$, and these are directed rays rather than an unspecified acute angle between two unoriented lines. From the sum and product of the parameters,

$$T_1-Q=(t_1-t_2)(t_1,1),\qquad T_2-Q=-(t_1-t_2)(t_2,1).$$

Taking their dot product and lengths gives

$$\cos\alpha=-\frac{p+1}{\sqrt{(t_1^2+1)(t_2^2+1)}}.$$

The denominator is positive. In particular, the sign of $p+1$ distinguishes acute and obtuse contact-ray angles. The determinant of the two ray vectors also gives

$$\sin\alpha=\frac{|t_1-t_2|}{\sqrt{(t_1^2+1)(t_2^2+1)}}.$$

Theorem 11.5.4

(Fixed contact-ray angle with its branch condition)

For a prescribed $0\lt\alpha\lt\pi$, $\alpha\ne\pi/2$, the exact locus of two-tangent intersections is

$$y^2-4x=(x+1)^2\tan^2\alpha,\qquad y^2-4x>0,$$

with $x\lt-1$ when the angle is acute, and $x>-1$ when it is obtuse. For $\alpha=\pi/2$, the exact locus is the directrix $x=-1$.

Dividing the sine and cosine expressions proves the squared equation when $p\ne-1$. Squaring loses the sign of the cosine, hence the branch condition must be retained. For sufficiency, a candidate with positive discriminant produces two distinct real contact parameters. Their squared tangent of the angle has the prescribed value, and the additional sign condition selects the correct one of the two supplementary angles. Thus the reconstructed rays have exactly the required angle, not merely an equal squared tangent.

For the perpendicular case the cosine formula gives $p=-1$. Conversely every point on this line has $\Delta=q^2+4>0$ and cosine zero, so the whole line is attained. The two points in the preceding example share the same parameter difference, but do not share the same contact-ray angle: $(-1,-1)$ gives a right angle, while $(5,5)$ gives an obtuse one.

Rolling externally: location and distance travelled

In the preceding loci, only the attainable positions mattered. A motion problem also asks when a point reaches a position and how far it has travelled. Keeping the parameter preserves that information and allows the path length to be computed from the derivative.

Let a circle of radius $b>0$ roll externally without slipping around a fixed circle of radius $a>0$. Initially the marked point is $(a,0)$, the point of contact. Let $\theta$ measure the orbital angle of the rolling centre. The line of centres turns through the same angle. Measure $\phi$ as the additional turning of the marked radius relative to this moving line, starting from its inward direction. The contact travels arc length $a\theta$ along the fixed circle; without slipping, the corresponding arc on the rolling circle has length $b\phi$. Hence

$$a\theta=b\phi.$$

The centre has position $(a+b)(\cos\theta,\sin\theta)$. The marked radius turns by $\theta$ with the line of centres and by a further $\phi$ relative to it. Its total turn in the fixed frame is therefore $\theta+\phi$; the minus sign below records that it initially points inward. Adding the centre position and this radius vector gives

$$P(\theta)=(a+b)(\cos\theta,\sin\theta) -b\left(\cos\frac{(a+b)\theta}{b},\sin\frac{(a+b)\theta}{b}\right).$$

The formula has the correct starting position. Keeping the radii positive and specifying forward motion also determines which parameter interval represents the requested part of the journey.

Worked example 11.5.5

(First return to the fixed circle when the radii have ratio two)

When $a=2b$, the position becomes

$$x=3b\cos\theta-b\cos3\theta,\qquad y=3b\sin\theta-b\sin3\theta.$$

Differentiate both coordinates. The squared speed with respect to this angular parameter is

$$(x')^2+(y')^2 =9b^2(2-2\cos2\theta)=36b^2\sin^2\theta.$$

Thus the speed is $6b|\sin\theta|$, with one power of the radius, as the units of length require. To locate the first return, compute the distance from the fixed centre:

$$|P(\theta)|^2=10b^2-6b^2\cos2\theta =4b^2+12b^2\sin^2\theta.$$

The marked point lies on the fixed circle exactly when $\sin\theta=0$. The smallest positive solution is $\theta=\pi$, where $P=(-2b,0)$, so this is the first return during forward motion. On $[0,\pi]$ the sine is nonnegative, and the distance travelled is

$$L=\int_0^\pi6b|\sin\theta|\,d\theta =6b[-\cos\theta]_0^\pi=12b.$$

This is the length of the travelled curve, not the straight-line distance between its endpoints. The speed vanishes at the contact endpoints, but the continuous nonnegative integrand still gives the arc length correctly.

A rational chart for the circle and ellipse

Trigonometric parameters describe the curve well, but a rational expression can make elimination easier. For two contacts, their parameter sum and product will then organize the tangent-intersection equations. A line through a fixed circle point supplies such a rational parameter.

The line $y=t(x+1)$ through $(-1,0)$ meets the unit circle at one further point. Solving the circle equation gives

$$\left(\frac{1-t^2}{1+t^2},\frac{2t}{1+t^2}\right),\qquad t\in\mathbb R.$$

The denominator is always positive. Conversely, every circle point other than $(-1,0)$ determines $t=y/(1+x)$, so this chart is one-to-one and covers exactly the circle minus that point. It is also the half-angle formula $t=\tan(\theta/2)$. Scaling gives the ellipse chart

$$\gamma(t)=\left(a\frac{1-t^2}{1+t^2},b\frac{2t}{1+t^2}\right), \qquad a>b>0,$$

which misses $(-a,0)$. A point missing from a chart is not missing from the ellipse; it must be handled separately when a construction uses it.

Worked example 11.5.6

(Rational derivation of the perpendicular-radius locus)

Take contacts $P=\gamma(t_1)$ and $Q=\gamma(t_2)$ with perpendicular radius vectors from the centre. Their tangents have a unique intersection: the contacts are neither equal nor antipodal. For finite chart parameters this means $t_1\ne t_2$ and $1+t_1t_2\ne0$.

Substitute the contact coordinates into the ellipse tangent equation and solve the two linear equations. Writing $r=t_1t_2$ and $s=t_1+t_2$ gives

$$x=a\frac{1-r}{1+r},\qquad y=b\frac{s}{1+r}.$$

For example, each tangent before solving is $(1-t_i^2)x/a+2t_i y/b=1+t_i^2$. The excluded zero denominator corresponds to antipodal contacts with parallel tangents, not to a finite intersection that may be recovered by cancellation.

Use a dot product for the perpendicular-radius condition, so axis contacts are included:

$$a^2(1-t_1^2)(1-t_2^2)+4b^2t_1t_2=0.$$

Now $(1-t_1^2)(1-t_2^2)=(1+r)^2-s^2$ and $4r=(1+r)^2-(1-r)^2$. Therefore

$$(a^2+b^2)(1+r)^2=b^2(1-r)^2+a^2s^2.$$

Divide by the nonzero squared denominator and use the intersection coordinates:

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

Section 11.2 proves the full converse by constructing normalized contact vectors from each candidate intersection. Here the rational calculation explains how the equation emerges through symmetric combinations of two parameters; it does not replace that existence argument.

Finally restore the missing chart contact. If $P=(-a,0)$, perpendicularity forces $Q=(0,\pm b)$. Their tangents meet at $(-a,\pm b)$, and both points satisfy the displayed locus equation. Thus chart omission creates no hole in the geometric locus once these cases are restored. The restriction concerns perpendicular radius vectors, not perpendicular tangent lines; the result is the established ellipse, not a director circle.

Common mistakes and summary

Elimination preserves equations more readily than inequalities. Keep the positive-hyperbola branch, the distinct-contact condition, and the sign of a ray-angle cosine visible. A missing rational-chart point and a genuinely impossible tangent intersection are different situations and need different repairs. For motion, use the first admissible parameter return and integrate speed rather than coordinate displacement.

These examples share a proof pattern: parameters encode the construction, symmetric relations simplify elimination, and the converse checks whether the resulting equation is the complete geometric answer.

Quick checks

Quick check

(Why does the midpoint locus need inequalities as well as an equation?)

Inspect the product of the two positive endpoint parameters.

Solution · Answer

The product is positive, so $x=1+st>1$ and $y=1+1/(st)>1$. The other branch of $(x-1)(y-1)=1$ cannot arise from endpoints on the positive branch.

Quick check

(Why can the squared angle equation describe the wrong contact-ray angle?)

Compare an angle with its supplement.

Solution · Answer

Supplementary angles have the same squared tangent but opposite cosine signs. The contact-ray cosine is proportional to $-(p+1)$, which selects the correct branch; the external-point condition guarantees two distinct contacts.

Exercises

1. For a positive-hyperbola chord through $(2,2)$ with $s=4$, find the other parameter and its midpoint. Check the restricted midpoint equation.
2. For $Q=(0,2)$, find both contact parameters and tangent equations to $y^2=4x$. Identify the tangent a finite-slope-only method would miss.
3. State the additional inequalities needed for the fixed contact-ray angle equation when the angle is acute, and when it is obtuse.
4. In the rolling-circle problem with $b=2$ and $a=4$, find the first-return parameter, return point, and distance travelled.
5. Explain which ellipse chart contact is omitted and recover its two perpendicular-radius tangent intersections.

Guided solutions

Solution · Model solution 1

$t=2/7$ and $M=(15/7,15/8)$. Its shifted-coordinate product is $(8/7)(7/8)=1$, and both coordinates exceed one.

Solution · Model solution 2

The equation is $t^2-2t=0$, so $t=0,2$. The tangents are $x=0$ and $x-2y+4=0$. The first is vertical and has no finite slope.

Solution · Model solution 3

Both require $y^2-4x>0$. For an acute angle also require $x\lt-1$; for an obtuse angle require $x>-1$. The perpendicular case is $x=-1$.

Solution · Model solution 4

The first return is at $\theta=\pi$, at $(-4,0)$, and the distance is $12b=24$.

Solution · Model solution 5

The omitted contact is $(-a,0)$. Its perpendicular radius contacts are $(0,\pm b)$; their tangents intersect at $(-a,\pm b)$.