FP1 4.4 - Loci problems with conics | Edexcel A-Level Further Mathematics

Find the path of a point built from a moving conic point or its tangents. You will eliminate parameters, prove which points are attainable and keep the restrictions that an equation alone can hide.

A locus needs an equation and its restrictions

Two distinct tangents to $y^2=8x$ have real parameters $t,u$ satisfying $tu=1$ . Their intersection stays on $x=2$ , yet it can never enter the segment between $(2,-4)$ and $(2,4)$ . Why can a moving point lie on a line but miss its middle? The real-parameter conditions decide the actual locus.

You already know

Use the conic parametrisations, focus-directrix properties, tangent equations and line-tangency conditions from FP1 4.1–4.3. The midpoint of $(x_1,y_1)$ and $(x_2,y_2)$ is $((x_1+x_2)/2,(y_1+y_2)/2)$ . Two finite perpendicular gradients have product $-1$ .

A locus is the set of all points satisfying a geometric rule. Use $(X,Y)$ for the moving point whose path is required, leaving $(x,y)$ or a parameter for the original conic point.

There are two directions to check. First show that the construction forces the claimed equation. Then show that each claimed point can actually arise, or state the restrictions that remove unattainable points. Squaring, cancelling factors and eliminating a parameter can all hide those restrictions.

For example, $X=t^2,\ Y=t^4$ , for real $t$ , gives $Y=X^2$ , but also $X\ge0$ . The negative- $X$ half of that parabola is not traced. The polynomial equation alone is too large a set.

Key idea

Write coordinates for the constructed point, eliminate the original parameter or line gradients, then check reality, distinctness, excluded values and the converse. State the final equation together with every restriction.

Check an especially simple domain restriction.

A midpoint transforms a conic

Start with a moving point $P=(4\cos t,3\sin t)$ on an ellipse and a fixed point $A=(h,k)$ . The midpoint $M=(X,Y)$ has

X=\frac h2+2\cos t,\qquad Y=\frac k2+\frac32\sin t

The cosine-sine identity eliminates $t$ . A complete parameter period is still important: it ensures that every point of the resulting ellipse is attained.

These transformed locus equations are derived results, not formulas supplied in the booklet.

[DIAGRAM: asset_name: The ellipse traced by a moving midpoint; asset_slug: 9fm0-fp173-midpoint-locus; caption: Taking a midpoint halves the semiaxes and moves the centre halfway towards the fixed point; description: Equal-scale coordinates show the original ellipse x squared over 16 plus y squared over 9 equals 1 in grey, the fixed point A at (2,0), and the midpoint locus centred at (1,0) with semiaxes 2 and 3 over 2 in blue. The original point P at (0,3) is joined to A, with midpoint M at (1,3 over 2) marked on the smaller ellipse. A separate note states the locus equation (X minus 1) squared over 4 plus 4Y squared over 9 equals 1.]
Taking a midpoint halves the semiaxes and moves the centre halfway towards the fixed point

Worked example 1: a fixed point to the right

Question. $P$ moves over $x^2/16+y^2/9=1$ . Find the locus of the midpoint of $P$ and $A=(2,0)$ .

Plan. Parametrise $P$ , write the midpoint coordinates and eliminate the parameter.

\begin{aligned} X&=1+2\cos t,\qquad Y=\frac32\sin t && \text{midpoint coordinates}\\ \frac{X-1}{2}&=\cos t,\qquad \frac{2Y}{3}=\sin t && \text{isolate the functions}\\ \frac{(X-1)^2}{4}+\frac{4Y^2}{9}&=1 && \text{sum of squares} \end{aligned}

Why this step? Subtract the new centre before squaring; replacing $(X-1)^2$ by $X^2-1$ would not describe a translated ellipse.

Answer. The complete ellipse $(X-1)^2/4+4Y^2/9=1$ .

Check. Conversely, from any point satisfying this equation, $P=(2X-2,2Y)$ lies on the original ellipse and has the required midpoint. No point of the new ellipse is missing.

Change only the fixed point's horizontal coordinate from $2$ to $-2$ .

Worked example 2: the fixed point to the left

Question. Find the midpoint locus when $A=(-2,0)$ , with the same moving $P$ .

Plan. Change the midpoint's centre while retaining the semiaxis scales.

\begin{aligned} X&=-1+2\cos t,\qquad Y=\frac32\sin t && \text{new fixed point}\\ \frac{(X+1)^2}{4}+\frac{4Y^2}{9}&=1 && \text{eliminate }t \end{aligned}

Why this step? The fixed point moves by $4$ units, but the midpoint centre moves by only $2$ ; the semiaxes remain half the original ones.

Answer. The complete ellipse $(X+1)^2/4+4Y^2/9=1$ .

Check. Recovering $P=(2X+2,2Y)$ from any locus point proves the converse.

This is an enlargement of scale factor $1/2$ about the fixed point, described through midpoint coordinates. The geometric construction explains both the new size and the new centre.

Watch out

Wrong: translate the centre but keep the original semiaxes. Right: the midpoint halves each coordinate's moving part, so it halves both semiaxes.

Move the fixed point vertically instead.

Tangent intersections and missing line segments

The tangents to $y^2=4ax$ at distinct parameters $t,u$ are $ty=x+at^2$ and $uy=x+au^2$ . If their intersection is $R=(X,Y)$ , subtracting gives

(t-u)Y=a(t-u)(t+u)

Since $t\ne u$ , cancellation is valid. Therefore

Y=a(t+u),\qquad X=atu

These intersection coordinates are not printed in the booklet; they follow from the tangent equations. The sum and product of $t,u$ also let us check whether real distinct contact parameters exist.

[DIAGRAM: asset_name: Complete and incomplete vertical loci of tangent intersections; asset_slug: 9fm0-fp173-tangent-loci; caption: Distinct real tangent parameters can remove a whole segment from a line; description: Two equal-scale coordinate panels use the parabola y squared equals 8x as light context. In the left panel the blue locus is x equals 2 with Y greater than 4 or less than minus 4; open circles at (2,4) and (2,minus 4) mark excluded endpoints and the middle segment is absent. In the right panel the sage locus is the whole line x equals minus 2, with no missing endpoint. Panel labels give tu equals 1 and tu equals minus 1. Arrowheads show that each locus extends beyond the plotting window.]
Distinct real tangent parameters can remove a whole segment from a line

Worked example 3: the opening missing segment

Question. Distinct real tangent parameters $t,u$ on $y^2=8x$ satisfy $tu=1$ . Find the exact locus of their intersection.

Plan. Use $a=2$ , then determine which real sums $t+u$ are possible with product $1$ .

\begin{aligned} X&=2tu=2,\qquad Y=2(t+u) && \text{intersection coordinates}\\ t,u>0&\ \Longrightarrow\ t+u\ge2,\qquad t,u<0\ \Longrightarrow\ t+u\le-2 && tu=1\\ Y&>4\quad\text{or}\quad Y<-4 && \text{equality would require }t=u \end{aligned}

Why this step? The endpoints come from repeated parameters $t=u=\pm1$ . They would describe a single tangent, rather than the intersection of two distinct tangents.

Answer. $X=2$ , with $Y>4$ or $Y<-4$ .

Check. Conversely, the required parameters are the roots of $z^2-(Y/2)z+1=0$ . Its discriminant is $Y^2/4-4>0$ exactly for the stated ranges, giving distinct real roots with product $1$ . Thus every stated point is attained.

Change only the product condition from $1$ to $-1$ . The parameters now have opposite signs.

Worked example 4: the whole line is attained

Question. Find the exact locus for distinct real tangent parameters satisfying $tu=-1$ on the same parabola.

Plan. Find the new fixed horizontal coordinate and test whether every real vertical coordinate is possible.

\begin{aligned} X&=2tu=-2,\qquad Y=2(t+u) && \text{intersection coordinates}\\ z^2-\frac Y2z-1&=0 && \text{required sum and product}\\ \Delta&=\frac{Y^2}{4}+4>0 && \text{every real }Y \end{aligned}

Why this step? A negative product guarantees opposite-sign, non-zero roots. They are automatically distinct, so no endpoint must be removed.

Answer. The complete line $X=-2$ , with $Y\in\mathbb R$ .

Check. $t=1,u=-1$ gives the midpoint of this line, $(-2,0)$ ; the positive discriminant proves that every other height is also possible.

The opening puzzle is resolved: elimination locates a line, while real distinct roots decide how much of that line belongs to the locus.

Watch out

Wrong: stop at $X=2$ in the positive-product case. Right: preserve the distinctness condition; it removes both endpoints as well as the entire middle segment.

Increase the positive product.

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FP1 4.4 - Loci problems with conics

A locus needs an equation and its restrictions

A midpoint transforms a conic

Tangent intersections and missing line segments

Continue this lesson.

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