Simplifying Calculations and Proofs with Parametric Equations

In one sentence

Write the quantity you want as a symmetric expression in some parameter, and substitute the whole block using Vieta’s formulas, without finding the intersection points.

The most useful parameter is the signed distance along a line. Write the line through $P_{0} (x_{0}, y_{0})$ with angle of inclination $α$ as

{x = x_{0} + t cos α y = y_{0} + t sin α

Since $(cos α, sin α)$ is a unit vector, $∣ t ∣$ is exactly the distance from the point to $P_{0}$ , and the sign of $t$ says which side of $P_{0}$ it is on. Substitute into a conic to get $a t^{2} + b t + c = 0$ ; its two roots $t_{1}, t_{2}$ are exactly $P_{0} A$ and $P_{0} B$ :

∣ P_{0} A ∣ \cdot ∣ P_{0} B ∣ = ∣ t_{1} t_{2} ∣ = \frac{c}{a}, midpoint ⟺ t_{1} + t_{2} = - \frac{b}{a} = 0, ∣ A B ∣ = ∣ t_{1} - t_{2} ∣.

Proving the intersecting chords theorem with the parameter t

Step 1 / 4

Write the line
Take the circle $x^{2} + y^{2} = r^{2}$ ( $r = 3$ in the picture) and any chord through a point $P_{0} (x_{0}, y_{0})$ inside it, with angle of inclination $α$ .

$x = x_{0} + t cos α, y = y_{0} + t sin α$
Substitute: a quadratic in t
Vieta’s formulas
This is the intersecting chords theorem

Predict first

$A B$ is a chord of the circle $x^{2} + y^{2} = 9$ through $P_{0} (1, 1)$ . Whatever its direction, what is $∣ P_{0} A ∣ \cdot ∣ P_{0} B ∣$ ?

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

Stop 1 · Ellipse: the chord with a given midpoint

$P_{0} (2, 1)$ is a point inside the ellipse $\frac{x ^{2}}{16} + \frac{y ^{2}}{4} = 1$ . Find the slope of the line containing the chord whose midpoint is $P_{0}$ .

Substitute the parametric line: the coefficient of $t$ is $2 (\frac{x _{0} cos α}{16} + \frac{y _{0} sin α}{4})$ . The midpoint condition is $t_{1} + t_{2} = 0$ , that is, this coefficient is $0$ :

\frac{2 cos α}{16} + \frac{sin α}{4} = 0 ⟹ tan α = - \frac{1}{2} .

Press “Turn to make P₀ the midpoint”, read off

α

and check that

tan α = - \frac{1}{2}

. Then drag

P_{0}

outside the ellipse: why is the button grayed out now?

In general, the chord with midpoint $(x_{0}, y_{0})$ has slope $- \frac{b ^{2} x _{0}}{a ^{2} y _{0}}$ , and we never found a single intersection point.

Stop 2 · Parabola: a focal chord

$F (1, 0)$ is the focus of $y^{2} = 4 x$ and $A B$ is a chord through $F$ . Prove that $∣ F A ∣ \cdot ∣ F B ∣ = ∣ A B ∣$ .

Substitute $x = 1 + t cos α, y = t sin α$ : $t^{2} sin^{2} α - 4 t cos α - 4 = 0$ .

∣ F A ∣ \cdot ∣ F B ∣ = ∣ t_{1} t_{2} ∣ = \frac{4}{sin ^{2} α}, ∣ A B ∣ = ∣ t_{1} - t_{2} ∣ = \frac{16 cos ^{2} α + 16 sin ^{2} α}{sin ^{2} α} = \frac{4}{sin ^{2} α} .

Turn

α

and compare the absolute value of “

t_{1} t_{2}

” with “

∣ A B ∣

”. This is the same fact as

\frac{1}{∣ F A ∣} + \frac{1}{∣ F B ∣} = 1

from §6.5 (polar coordinates): multiply both sides by

∣ F A ∣ \cdot ∣ F B ∣

Stop 3 · An angle parameter: a two-variable extremum becomes one-variable trigonometry

One problem, several solutions

The point $P (x, y)$ lies on the ellipse $\frac{x ^{2}}{16} + \frac{y ^{2}}{4} = 1$ . Find the maximum of $x + 2 y$ .

Set x + 2y = m and use the discriminant

Substitute $x = m - 2 y$ : $(m - 2 y)^{2} + 4 y^{2} = 16$ , that is $8 y^{2} - 4 m y + m^{2} - 16 = 0$ . For $y$ to have a real solution, $Δ = 16 m^{2} - 32 (m^{2} - 16) \geq 0$ , so $m^{2} \leq 32$ and the maximum is $42$ .

CostSubstitute, tidy up, solve an inequality: three steps, each easy to get wrong.

The eccentric angle as parameter

Set $x = 4 cos φ, y = 2 sin φ$ . Then $x + 2 y = 4 cos φ + 4 sin φ = 42 sin (φ + \frac{π}{4}) \leq 42$ , reached at $φ = \frac{π}{4}$ , that is at $P (22, 2)$ .

CostOne line, and you see at once where the maximum is reached.

The parameter turns “two variables + one constraint” into “one variable, no constraint”. The extremum problems at the next stop use this move again and again.

Practice

Exercise 1Basic

$P_{0} (3, 2)$ is a point inside the circle $x^{2} + y^{2} = 25$ . Any chord through $P_{0}$ is cut by $P_{0}$ into two pieces. Find the product of their lengths.

Exercise 2Basic

For the ellipse $\frac{x ^{2}}{9} + \frac{y ^{2}}{4} = 1$ , what is the slope of the line containing the chord whose midpoint is $P_{0} (1, 1)$ ?

Exercise 3Advanced

The line through $(2, 0)$ with angle of inclination $\frac{π}{4}$ meets the parabola $y^{2} = 4 x$ at $A$ and $B$ . Find $∣ A B ∣^{2}$ .

Exercise 4Challenge

$P_{0} (5, 0)$ is a point outside the circle $x^{2} + y^{2} = 9$ . A secant through $P_{0}$ meets the circle at $A$ and $B$ with $∣ P_{0} A ∣ = 2$ ( $A$ is the nearer one). Find $∣ A B ∣$ .

In one sentence

Proving the intersecting chords theorem with the parameter t

Stop 1 · Ellipse: the chord with a given midpoint

Stop 2 · Parabola: a focal chord

Stop 3 · An angle parameter: a two-variable extremum becomes one-variable trigonometry

Set x + 2y = m and use the discriminant

The eccentric angle as parameter

Practice

Mistake Museum · Exhibits from this stop

Finding the “chord with a given midpoint” from $t_1+t_2=0$ without checking that the chord exists

Taking $t_1t_2$ to be $|P_0A|\cdot|P_0B|$ directly