1. Parametric Equations I
x2 = 4ay
Tangent
Chord
Normal
P(2ap, ap2)
By Mr Porter

2. The Parametric Equation of the Chord.
Find the equation of the chord joining P(2ap, ap2) and Q(2aq, aq2), two distinct points on the parabola x2 = 4ay. Show that if PQ is a focal chord, then pq = -1.
Q(2aq, aq2)
P(2ap, ap2)
x2 = 4ay
Equation of Chord PQ:
P(2ap, ap2)
Partially expand
Make ‘y’ the subject
Gradient of Chord PQ
If PQ is a focal chord, then S(0, a)
lies on the chord PQ.
Factorise and simplify
Divide by ‘-a’
for PQ to be a focal chord.

3. General Tangent and Normal.
Find the equation of the tangent and normal at the point P(2at, at2) to the parabola x2 = 4ay.
x2 = 4ay
Tangent
Normal
P(2at, at2)
equation of the tangent:
gradient of normal:
equation of the normal:
Gradient of tangent: at x = 2at.

4. Example 1
Show that the tangent to x2 = 12y at the parametric point P(6p, 3p2) has equation y = px – 3p2.
By substituting the point A(2, -1) into the equation of the tangent, find the cartesian points of contact, and the cartesian equations, of the tangents to the parabola from A.
equation of the tangent:
x2 = 12y
P(6p, 3p2)
A(1, -2) Q
b) substituting the point A(2, -1) into tangent
a) Gradient of tangent: at x = 6p.
cartesian points of contact (6p, 3p2) are
cartesian equations, of the tangents y = px – 3p2 are
y = x – 3 and

5. Example 2A
P(2ap, ap2), Q(2aq, aq2) are two points on the parabola x2 = 4ay. Show that the point T of intersection of the tangents at P, Q is given by T = [a(p + q), apq].
P(2ap, ap2)
Q(2aq, aq2)
x2 = 4ay T
Likewise, tangent at Q: y = qx – aq2
T, point of intersection, solve the tangent equations simultaneously.
y = px – ap2 and y = qx – aq2
px – ap2 = qx – aq2
px – qx + aq2 – ap2 = 0
Factorise.
Grad of tangent at P:
(p – q)x – a(p2 – q2) = 0
(p – q)x – a(p – q)(p + q) = 0
Divide by (p – q)
x – a(p + q) = 0
x = a(p + q)
equation of the tangent:
Substitute x into y = px – ap2
y = pa(p + q) – ap2
y = apq
Hence, T (x, y) = [a(p + q), apq]

6. Example 2B
Tangents drawn from two variable points P(2ap, ap2) and Q(2aq, aq2) on the parabola x2 = 4ay intersection at right angles at point T. Find the cartesian equation of the locus of T.
T, point of intersection, solve the tangent equations simultaneously.
y = px – ap2 and y = qx – aq2
px – ap2 = qx – aq2
px – qx + aq2 – ap2 = 0
Factorise.
(p – q)x – a(p2 – q2) = 0
(p – q)x – a(p – q)(p + q) = 0
x – a(p + q) = 0
x = a(p + q)
Substitute x into y = px – ap2
y = pa(p + q) – ap2
y = apq, but pq = -1
y = -a
Hence, T (x, y) = [a(p + q), -a]
Divide by (p – q)
Grad of tangent at P:
equation of the tangent:
Likewise, tangent at Q:
y = qx – aq2
From Example 2A, the equations of tangents
at P and Q are
P(2ap, ap2)
Q(2aq, aq2)
x2 = 4ay T
and
From Example 2A, the point of intersection T is
Grad of tangent at P:
T (x, y) = [a(p + q), apq]
Then, T is
Cartesian equation of the locus of T is:
y = -a
Note: y = -a is the equation of the directrix.
Therefore, grad of tangent at Q
mq = q
but mq x mp = -1
Student must be able to derive the equations of the tangents at P and Q. Also, students need to show how to find the point of intersection, T.

7. Example 2C
Prove that the tangents to the parabola, x2 = 4ay, at the extremities of a focal chord intersect at right angles on the directrix.
Equation of the tangent:
Substitute x into y = px – ap2
y = pa(p + q) – ap2
Let the coordinates of the extremities of the focal chord be P(2ap, ap2) and Q(2aq, aq2) on the parabola x2 = 4ay.
If PQ is a focal chord, then S(0, a)
lies on the chord PQ.
y = apq, but pq = -1
y = -a
Divide by ‘-a’
Gradient of Chord PQ
Likewise, tangent at Q:
y = qx – aq2
Hence, T (x, y) = [a(p + q), -a]
y = -a is the equation of the directrix. Therefore, T lies on the directrix.
for PQ to be a focal chord.
Gradient of tangent: at P .
T, point of intersection, solve the tangent equations simultaneously.
mP = p, likewise mQ = q
y = px – ap2 and y = qx – aq2
px – ap2 = qx – aq2
Equation of Chord PQ:
px – qx + aq2 – ap2 = 0
Factorise.
(p – q)x – a(p2 – q2) = 0
(p – q)x – a(p – q)(p + q) = 0
x – a(p + q) = 0
Divide by (p – q)
P(2ap, ap2)
For the tangents to be perpendicular
mP x mQ = -1
But, p x q = -1, condition to be a focal chord.
Tangent intersect at right angles.
x = a(p + q)

8. Example 3
The normal at any point P(4p, -2p2) on the parabola cuts the Y-axis at T.
Find the cartesian equation of the locus of the mid-point Q of PT.
locus of the point Q
Gradient of tangent:
Equation of normal:
at T, x = 0
Therefore, T = (0, -4 – 2p2)
Q, the mid-point of PT is:
i.e. (x – h)2 = –4a( y – k) a parabola
Vertex (h, k) = (0, -2)
Focal length a = ½
Gradient of normal: