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

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.

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.

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

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]

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.

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)

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: