Parametrics Equations Proving (1 Viewer)

Riesla · Nov 1, 2016

5. PQ is a focal chord in the parabola x²=4ay

If PT is drawn parallel to the tangent at Q and QT is drawn parallel to the tangent at P, show that the locus of T is x²=a(y-3a)

Help me solve the following please thanks :>

Mahan1 · Nov 18, 2016

Riesla said:
5. PQ is a focal chord in the parabola x²=4ay

If PT is drawn parallel to the tangent at Q and QT is drawn parallel to the tangent at P, show that the locus of T is x²=a(y-3a)

Help me solve the following please thanks :>

This questions involves some presumed knowledge(You can easily derive them)
First: if PQ is a focal length then pq =-1 (Find the question passing through P and Q and use the fact that it passes through focal point (0,a) )
and the slope of tangent at P and Q is p and q respectively. (differentiate $\frac{y}{4a} = x^2$ with respect to x and sub in the parametric form of x and y at points P and Q)
Knowing that we get :

line QT : $y = px + b$
since QT passes through Q then
$aq^{2} = p(2aq) + b \Rightarrow b = aq^{2} - 2apq = aq^{2} + 2a$

similarly line PT : $y = qx + b$
therefore
$b = ap^{2} -2apq = 2a+ap^{2}$ . by equating those equations we obtain:

$y = px+aq^{2} + 2a = qx + ap^{2} + 2a$
that implies
$x = a(p+q), y = ap^{2} + aq^{2} + a$
then
$a(y-3a) = a(ap^{2} + aq^{2} a-3a) = a(ap^{2} + aq^{2} -2a) = a^{2} (p^{2} + q^{2} -2) = (a(p+q))^{2} = x^{2}$
(since $(q+p)^{2}= p^{2} + q^{2} + 2pq = p^{2} + q^{2} -2)$

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Parametrics Equations Proving (1 Viewer)

Riesla

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