Locus

Question 0

Simple locus circle

Answer 0

A circle is the locus of all points a fixed distance (radius) from a point (centre)
Sample Question:
Find the equation of the locus of a point P(x,y) that moves so that it is always 5 units away from (1,0)

Question 1

A locus is:

Answer 1

a) set of points which satisfy a given condition (x,y)
b) point P(x,y) that moves along a certain path
A locus is an equation which describes the points

Question 2

Simple locus SLG

Answer 2

A SLG is the locus of points equidistant from two points or lines.

Question 3

Simple locus parabola

Answer 3

The locus of all points equidistant from a point and a line

Question 4

Parabolas locus in basic form

Answer 4

x^2= +/- 4ay

Question 5

Focus=

Answer 5

F(0,a)

Question 6

Latus rectum=

Answer 6

4a

Question 7

Directrix=

Answer 7

y= +/- a

Question 8

Parabolas in inverse basic form

Answer 8

y^2= +/- 4ax

Question 9

Directrix in inverse form=

Answer 9

x=+/- a

Question 10

Inverse focus=

Answer 10

F(a,0)

Question 11

Parabolas in general form

Answer 11

(x-A)^2=4a(y-B) where the vertex is (A,B)

Question 12

General form focus=

Answer 12

F(A,B+a)

Question 13

General form directrix=

Answer 13

y=B-a

Question 14

Parabolas in inverse general form

Answer 14

(y-B)^2= 4a(x-A)
IMPORTANT
Notice how not only the x and y values have replaced each other, but the whole (y-B) and (x-A) terms.

Question 15

Parabolas in inverse general form focus=

Answer 15

F(A+a, B)

Question 16

Parabolas in inverse general form directrix=

Answer 16

x=A-a

Question 17

How to convert a parabola equation in general form locus to find its vertex:

Answer 17

1) locate the term with the highest power
2) move all terms that do not contain the term with the highest power to the RHS
E.g. x^2-6x-16y=41 —> x^2-6x=16y+41
3) complete the square for the LHS
4) factorise the RHS and change it into +/-4a(y-B) form

Question 18

Parametric Equations

Answer 18

–> equations in which two parameters are expressed in terms of a third parameter i.e. X and y in terms of t

Question 19

Parametric equations of the parabola:

Answer 19

–>the cartesian equation of x^2=4ay can be re-expressed in the parametric form:
x=2at
y=at^2

Question 20

Converting parametric to cartesian:

Answer 20

1) set up a simultaneous equation
2) from (1) make t the subject
3) sub t into (2) and simplify

Question 21

Converting cartesian to parametric

Answer 21

1) Express the cartesian equation in x^2 = 4ay to find a

2) sub a into each line of x=2at and y=at^2

Question 22

Locus and the Parabola Rules

Rule 1
Coordinates

Answer 22

Any point, P, on the parabola x^2=4ay, has the coordinates, P(2ap, ap^2)

Question 23

Locus and the Parabola Rules

Rule 2
Chord

Answer 23

If P(2ap, ap^2) and Q(2aq, aq^2) are any two points on the parabola x^2=4ay, then the chord PQ has:
i) gradient: m= p+q/2
ii) equation: y-1/2(p+q)x+apq=0
Proof:
i) gradient formula with P and Q
ii) Point gradient formula

Question 24

Locus and the Parabola Rules

Rule 3
Focal Chord

Answer 24

If PQ is a focal chord then pq=-1
Proof:
Sub focus (0,a) into the PQ equation

Question 25

Locus and the Parabola Rules

Rule 4
Tangent

Answer 25

The tangent to the parabola x^2=4ay at the point P(2ap, ap^2) has
i) gradient: m=p
ii) equation: y-px+ap^2=0
Proof
i) derivative
ii) point gradient formula

Question 26

Locus and the Parabola Rules

Rule 5
Tangent Intersection

Answer 26

The tangents to the parabola x^2=4ay at the points P(2ap, ap^2) and Q(2aq, aq^2) intersect at the point (a(p+q),apq)
Proof:
2 tangents simultaneous equation

Question 27

Locus and the Parabola Rules

Rule 6
Normal

Answer 27

The normal to the parabola x^2=4ay at the point P(2ap, ap^2) has
i) gradient: m= -1/p
ii) equation: x+py= ap^3+2ap
Proof
i) m1m2=-1 to tangent gradient
ii) point gradient formula

Question 28

Locus and the Parabola Rules

Rule 7
Normal Intersection

Answer 28

The normals to the parabola x^2=4ay at the points P(2ap, ap^2) and Q(2aq, aq^2) intersect at the point
(-apq(p+q), a(p^2+pq+q^2+2))
Proof
Simultaneous equations of normals

Question 29

Locus and the Parabola Rules

Rule 8
Cartesian Tangent

Answer 29

If point A(x1, y1) lies on the parabola x^2=4ay then the equation of the tangent at A is given by
                                     x x1=2a(y+y1)
Proof
Gradient (derivative) then point gradient formula

Question 30

Locus and the Parabola Rules

Rule 9
Cartesian Normal

Answer 30

If point A(x1,y1) lies on the parabola x^2=4ay then the normal at A is given by
y-y1=-2a/x1(x-x1)
Proof
Derivative, m1m2=-1 of tangent then point gradient formula

Question 31

Locus and the Parabola Rules

Rule 10
Chord of Contact

Answer 31

The equation of the chord of contact, XY, of tangents drawn from the point P(x1, y1) to the parabola x^2=4ay is given by
x x1= 2a(y+y1)
Proof
Equation of chord XY
Intersection of tangents equal to P
x1=a(p+q)—1 y1=apq—2
Find (p+q), apq to simplify the equation of the chord