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Conics · Conics

Conic sections and loci

A telecommunications aerial, the orbit of a planet, the cable of a suspension bridge and the reflector of a lighthouse have nothing to do with one another. And yet the curves they describe are only four, and all four come from cutting a cone with a plane. You are going to meet them as loci: the set of points that satisfy a condition on distances. You already know the circle from 3rd ESO; now come the ellipse, the hyperbola and the parabola.

Before you start, make sure you know: Distance between two points (4th ESO)
Before you start, make sure you know: Special products (algebraic identities) (2nd ESO)

Cutting a cone: where conics come from

Conic sections obtained by cutting cones: a plane perpendicular to the axis produces a circle, a slanted plane produces an ellipse and a plane parallel to a generator produces a parabola. circle (plane ⟂ to the axis) ellipse (slanted plane) parabola (plane ∥ to a generator)
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A cone of revolution is cut by a plane that does not pass through its vertex. If the plane is parallel to a single generator of the cone, what conic section is obtained?…

A cone of revolution is cut by a plane that does not pass through its vertex. If the plane is parallel to a single generator of the cone, what conic section is obtained? Justify your answer based on the geometric definition of conic sections.

Solve

Revision: the locus as a tool

Revise the circle as a locus: drag P until $OP$ is always the same length.

The four conics in this topic are slightly more elaborate loci: instead of a single fixed distance, they impose a relationship between two distances (to one or two fixed points, or to a point and a line).

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Find the equation of the locus of points in the plane whose distance from the origin of coordinates is always equal to $5$.

Find the equation of the locus of points in the plane whose distance from the origin of coordinates is always equal to $5$.

Solve
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Find the equation of the locus of points in the plane that are equidistant from points $A(2, -3)$ and $B(-4, 5)$. What geometric shape does it represent?

Find the equation of the locus of points in the plane that are equidistant from points $A(2, -3)$ and $B(-4, 5)$. What geometric shape does it represent?

Solve

Building the equation step by step

The method is always the same: you call the general point $P(x,y)$, write the condition using the distance formulae, and simplify until the roots disappear.

Three-step diagram for finding the equation of a locus: set up the condition using distances, square to remove the roots and simplify the quadratic terms. condition withd(P,A), d(P,B) √((x+3)²+y²) = 2√((x−3)²+y²) square both sides (x+3)²+y² = 4[(x−3)²+y²] simplify (x−5)²+y² = 16
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In a telecommunications project, two transmitting antennas are located at points $A(-3,0)$ and $B(3,0)$ on a Cartesian plane (in km). A receiver $R(x,y)$ moves such that…

In a telecommunications project, two transmitting antennas are located at points $A(-3,0)$ and $B(3,0)$ on a Cartesian plane (in km). A receiver $R(x,y)$ moves such that its distance to antenna $A$ is always twice its distance to antenna $B$. Find the equation of the locus described by the receiver and identify which conic section it is, stating its centre and radius.

Solve
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The end $P(x,y)$ of an industrial robotic arm is programmed to move such that its distance to the safety line $x = -4$ is always twice its distance to the anchor point…

The end $P(x,y)$ of an industrial robotic arm is programmed to move such that its distance to the safety line $x = -4$ is always twice its distance to the anchor point $F(-1,0)$. Determine the equation of the path described by the end of the arm and identify what type of conic section it represents.

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Conditions with squared distances

Some conditions come without roots already: they ask for the sum or the difference of the squares of two distances. They are even quicker to work with.

Diagram of two fixed points A and B and a moving point P(x,y), showing that the sum of squared distances defines a circle and their difference defines a straight line. A B P(x,y) d(P,A)² + d(P,B)² = const → circle d(P,A)² − d(P,B)² = const → line
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Find the locus of points $P(x,y)$ whose difference of squared distances to points $A(1,0)$ and $B(-1,0)$ is equal to $8$, that is, $d(P,A)^2 - d(P,B)^2 = 8$.

Find the locus of points $P(x,y)$ whose difference of squared distances to points $A(1,0)$ and $B(-1,0)$ is equal to $8$, that is, $d(P,A)^2 - d(P,B)^2 = 8$.

Solve
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Find the equation of the locus of points $P(x,y)$ such that the sum of the squares of their distances to points $A(2,0)$ and $B(-2,0)$ is equal to $20$.

Find the equation of the locus of points $P(x,y)$ such that the sum of the squares of their distances to points $A(2,0)$ and $B(-2,0)$ is equal to $20$.

Solve

Eccentricity: one recipe for all four curves

This number $e$ —the eccentricity— will reappear in each of the next three sections: it measures how "flattened" the curve is compared with its simplest form, the circle.

Classification of the conics by eccentricity: circle with $e = 0$, ellipse with $0 ‹ e ‹ 1$, parabola with $e = 1$ and hyperbola with $e › 1$ shown on a half-line. circleellipseparabolahyperbola e = 00 < e < 1e = 1e > 1
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The reflector of a lighthouse has a profile designed such that the distance from any point on its surface to the light-emitting focus, situated at $F(0,2)$, is exactly…

The reflector of a lighthouse has a profile designed such that the distance from any point on its surface to the light-emitting focus, situated at $F(0,2)$, is exactly equal to its distance from a horizontal reference line with equation $y = -2$. Determine the equation of the locus of the reflector's profile.

Solve

Back to the beginning: each application uses a different curve: the orbits of the planets are ellipses; the reflectors of lighthouses and the cables of bridges take parabolic shapes; and the synchronisation of signals between aerials relies on the hyperbola and the parabola.

Summary

See all the exercises in this section