Methods for Locus Problems

In one sentence

To find a locus, first look at what drives the moving point. How it is driven decides the method:

How the point moves	Method	What to do
The condition itself is a geometric relation for $P$	Direct method	Let $P (x, y)$ and translate the condition word for word into coordinates
It follows another point $P_{0}$ on a known curve	Substitution (related-point) method	Express $P_{0}$ in terms of $(x, y)$ and substitute into the equation $P_{0}$ satisfies
It is driven by an angle, a slope, time…	Parameter method	Write $x = f (t), y = g (t)$ , then eliminate $t$
It is where two moving lines (or curves) meet	Intersecting-loci method	Write the equations of the two lines and eliminate their common parameter directly

Every method ends by going back to the previous stop: check purity and completeness.

Substitution method: P is on the circle x² + y² = 16, Q is the midpoint of P and A(6, 0)

Step 1 / 5

See what is driving
$Q$ has no condition of its own; it simply follows the point $P$ on the circle. $P$ satisfies the known equation $x^{2} + y^{2} = 16$ .

$P is on x^{2} + y^{2} = 16$
Name the point you want
Express P the other way round
Substitute into the known equation
Check

Predict first

$A (- 3, 0)$ , $B (3, 0)$ . What is the locus of points $P$ with $∣ P A ∣ = 2∣ P B ∣$ ?

The perpendicular bisector of

A B

A circleAn ellipseOne branch of a hyperbola

Choose an answer first; then you can see the result.

Stop 1 · Parameter method: a midpoint driven by the angle of inclination

Chords of the parabola $y^{2} = 4 x$ are drawn through its focus $F (1, 0)$ . Find the locus of the chord midpoint $M$ .

Why is there no point when the angle of inclination is

α = 0

? (That line is the axis of the parabola.)

One problem, several solutions

$A B$ is a chord of the parabola $y^{2} = 4 x$ through its focus $F (1, 0)$ . Find the equation of the locus of its midpoint $M$ .

Parameter method: use the slope k

Substitute the chord $y = k (x - 1)$ : $k^{2} x^{2} - (2 k^{2} + 4) x + k^{2} = 0$ , so $x_{1} + x_{2} = 2 + \frac{4}{k ^{2}}$ . The midpoint has $x = 1 + \frac{2}{k ^{2}}$ and $y = k (x - 1) = \frac{2}{k}$ . Eliminating $k$ : $y^{2} = \frac{4}{k ^{2}} = 2 (x - 1)$ . The vertical chord $x = 1$ has midpoint $F (1, 0)$ , which also satisfies the equation. The locus is $y^{2} = 2 (x - 1)$ .

CostStraightforward: one elimination, plus the vertical chord handled separately.

Subtracting the endpoint equations: set up, never solve

$y_{1}^{2} = 4 x_{1}$ and $y_{2}^{2} = 4 x_{2}$ ; subtracting, $(y_{1} + y_{2}) (y_{1} - y_{2}) = 4 (x_{1} - x_{2})$ . Let $M (x, y)$ . Then $y_{1} + y_{2} = 2 y$ , and the slope of the chord is $\frac{y _{1} - y _{2}}{x _{1} - x _{2}} = \frac{y}{x - 1}$ (both $M$ and $F$ are on the chord). Substituting: $2 y \cdot \frac{y}{x - 1} = 4$ , that is $y^{2} = 2 (x - 1)$ .

CostShorter, but you must remember two assumptions: the slope exists, and the line through M and F is the chord.

Stop 2 · Intersecting-loci method: where two moving lines meet

Through $A (- 2, 0)$ draw the line with slope $k$ , and through $B (2, 0)$ the line with slope $\frac{1}{k}$ . Find the locus of their intersection $P$ .

The two lines are $y = k (x + 2)$ and $y = \frac{1}{k} (x - 2)$ . There is no need to solve for the intersection: multiply the two equations, and $k$ times $\frac{1}{k}$ simply disappears:

y^{2} = (x + 2) (x - 2) ⟹ x^{2} - y^{2} = 4 .

Drag

θ

from

- 8 9^{\circ}

8 9^{\circ}

. Where does the intersection point go at

θ = \pm 4 5^{\circ}

? As

θ \to 0

, which point does it approach without ever reaching it?

The heart of the intersecting-loci method is “eliminate the parameter, never find the intersection”: whether you multiply, divide, add or subtract depends on which combination makes the parameter vanish by itself.

Practice

Exercise 1Basic

$A (- 2, 0)$ , $B (2, 0)$ , and the point $P$ satisfies $∣ P A ∣^{2} + ∣ P B ∣^{2} = 26$ . The locus of $P$ is a circle. Find its radius.

Exercise 2Basic

The point $P$ moves on the parabola $y = x^{2}$ , and $Q$ is the midpoint of the segment $O P$ ( $O$ is the origin). Write the equation of the locus of $Q$ as $y = F (x)$ . What is $F (x)$ ?

Exercise 3Advanced

A moving line $y = k x$ through the origin meets the circle $(x - 2)^{2} + y^{2} = 4$ at the origin and at another point $M$ . The midpoint $N$ of $O M$ moves on a circle. Find the radius of that circle.

Exercise 4Challenge

$A (- 3, 0)$ , $B (3, 0)$ . Lines are drawn through $A$ and through $B$ whose slopes have product $- \frac{4}{9}$ ; they meet at $P$ . The locus of $P$ is the ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ with two points removed. Find $a^{2} + b^{2}$ .

In one sentence

Substitution method: P is on the circle x² + y² = 16, Q is the midpoint of P and A(6, 0)

Stop 1 · Parameter method: a midpoint driven by the angle of inclination

Parameter method: use the slope k

Subtracting the endpoint equations: set up, never solve

Stop 2 · Intersecting-loci method: where two moving lines meet

Practice

Mistake Museum · Exhibits from this stop

Substituting the wrong way round in the substitution method

Forgetting the excluded points after eliminating the parameter in the intersecting-loci method