Quadratics - the Australian Mathematical Sciences Institute
Quadratics - the Australian Mathematical Sciences Institute
Quadratics - the Australian Mathematical Sciences Institute
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A guide for teachers – Years 11 and 12 • {29}<br />
Example<br />
Suppose P(2ap, ap 2 ) and Q(2aq, aq 2 ) are two points lying on <strong>the</strong> parabola x 2 = 4ay.<br />
Prove that, if OP and OQ are perpendicular, <strong>the</strong>n pq = −4.<br />
Solution<br />
The gradient of OP is ap2<br />
2ap = p 2 and similarly <strong>the</strong> gradient of OQ is q . Since <strong>the</strong>se lines<br />
2<br />
are perpendicular, <strong>the</strong> product of <strong>the</strong>ir gradients is −1 and so p 2 × q = −1, which implies<br />
2<br />
that pq = −4.<br />
Note that, in this case, PQ can never be a focal chord.<br />
Exercise 15<br />
Suppose P(2ap, ap 2 ) and Q(2aq, aq 2 ) are two points lying on <strong>the</strong> parabola x 2 = 4ay.<br />
Prove that, if PQ is a focal chord, <strong>the</strong>n <strong>the</strong> length of <strong>the</strong> chord PQ is given by a ( p + 1 p<br />
) 2.<br />
Tangents<br />
We have seen that <strong>the</strong> gradient of <strong>the</strong> chord joining two distinct points P(2ap, ap 2 ) and<br />
Q(2aq, aq 2 ) on <strong>the</strong> parabola x 2 = 4ay is 1 2<br />
(p + q).<br />
If we now imagine <strong>the</strong> point Q moving along <strong>the</strong> parabola to <strong>the</strong> point P, <strong>the</strong>n <strong>the</strong> line<br />
through P and Q will become a tangent to <strong>the</strong> parabola at <strong>the</strong> point P. Putting q = p, we<br />
see that <strong>the</strong> gradient of this tangent is 1 2<br />
(p + p) = p.<br />
x<br />
x2 = 4ay<br />
y<br />
P(2ap‚ap2)<br />
0<br />
A tangent to <strong>the</strong> parabola.