The point ${\rm{T}}(a{t^2},2at)$ lies on the parabola ${y^2} = 4ax$ . Show that the tangent to the parabola at T has equation $y = \frac{x}{t} + at$ .

The distinct points ${\rm{P}}(a{p^2}, 2ap)$ and $Q(a{q^2}, 2aq)$ , where $p$, $q \ne 0$ , also lie on the parabola ${y^2} = 4ax$ . Given that the line (PQ) passes through the focus, show that

  (i)     $pq = - 1$ ;

  (ii)     the tangents to the parabola at P and Q, intersect on the directrix.

$2y\frac{{{\rm{d}}y}}{{{\rm{d}}x}} = 4a$     M1

$\frac{{{\rm{d}}y}}{{{\rm{d}}x}} = \frac{{2a}}{y} = \frac{1}{t}$     A1

Note: Accept parametric differentiation.

the equation of the tangent is

$y - 2at = \frac{1}{t}(x - a{t^2})$     A1

$y = \frac{x}{t} + at$     AG

Note: Accept equivalent based on $y = mx + c$ .

[3 marks]

(i)     the focus F is ($a$, $0$)     A1

EITHER

the gradient of (PQ) is $\frac{{2a(p - q)}}{{a({p^2} - {q^2})}} = \frac{2}{{p + q}}$     M1A1

the equation of (PQ) is $y = \frac{{2x}}{{p + q}} + \frac{{2apq}}{{p + q}}$     A1

substitute $x = a$ , $y = 0$     M1

$pq = - 1$     AG

OR

the condition for PFQ to be collinear is

$\frac{{2a(p - q)}}{{a({p^2} - {q^2})}} = \frac{{2ap}}{{a{p^2} - a}}$     M1A1

$\frac{2}{{p + q}} = \frac{{2p}}{{{p^2} - 1}}$     A1

${p^2} - 1 = {p^2} + pq$     A1

$pq = - 1$     AG

Note: There are alternative approaches.

(ii)     the equations of the tangents at P and Q are

$y = \frac{x}{p} + ap$ and $y = \frac{x}{q} + aq$

the tangents meet where

$\frac{x}{p} + ap = \frac{x}{q} + aq$     M1

$x = apq = - a$     A1

the equation of the directrix is $x = - a$     R1

so that the tangents meet on the directrix     AG

[8 marks]