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]