forgot something simple... (1 Viewer)

[sasquatch]

Two two points P(2ap, ap²) and Q(2aq, aq²) are on the parabola x² = 4ay.

i) The equation of the tangent to x² = 4ay at an arbitary points (2at, at²) on the parabola is y = tx - at².

Show that the tangents at the points P and Q meet at R, where R is the point (a(p+q),apq).

** I can do part i, but get stuck at part ii **

ii) As P varies the point Q is always chosen so that /_POQ is a right angle, where O is the origin. Find the locus of R.



What i have done is the following (excluding part i)

x = a(p+q)
y = apq

x² = a²(p² + 2pq + q²)
x² = a(ap² + 2apq + aq²)
x² = a(ap² + aq² + 2y)

I cannot get rid of the ap² + aq²... I cant remember if this is alright or.....well yeah i forget. I cant even remember if i did it the right way.

 

[Trev]

Remember since angle POQ is a right angle that m_OPm_OQ=-1.
Differentiating x²=4ay and finding gradients it works out that pq=-1.

So,
ap²+aq² = a(p²+q²) = a[(p+q)²-2pq] = a[(x/a)²+2].

 

[sasquatch]

um i think pq = -4.. cuz its not the tangents... its OP and OQ which are

m_OP = ap<sup>2[/sup} / 2ap
m_OQ = aq2[/sup} / 2aq

m_OP * m_OQ = -1
pq/ 4 = -1
pq = -4

ap²+aq² = a(p²+q²) = a[(p+q)²-2pq] = a[(x/a)²+8].

but if you do it like that....... and you put it into

x2 = a(a[(x/a)2+8] + 2y)
x2 = a[(x2/a)+8] + 2y)
x2 = ax2/a +8a + 2ay
x2 = 2 + 8a + 2ay
2ay = -8a
y = -4

is that right then?</sup>

 

[Riviet]

pq=-4 is correct, I think you need to find an expression for p+q=? to be able to substitute into a(p+q)² so you get the locus in cartesian form.

 

[sasquatch]

Yeah i was thinking what he did was a little weird.. cuz if you were to do

x = a(p+q)
y = apq

x² = a²(p² + 2pq + q²)
x² =a²[(p + q)²)
x² = a²(x²/a²)
x² = x²

which obviously gets you no where.... can someone try doing the question please?

 

[Trev]

Eh, same thing.

 

[Riviet]

um i think pq = -4.. cuz its not the tangents... its OP and OQ which are

m_OP = ap<sup>2[/sup} / 2ap
m_OQ = aq2[/sup} / 2aq

m_OP * m_OQ = -1
pq/ 4 = -1
pq = -4

ap²+aq² = a(p²+q²) = a[(p+q)²-2pq] = a[(x/a)²+8].

but if you do it like that....... and you put it into

x2 = a(a[(x/a)2+8] + 2y)
x2 = a[(x2/a)+8a + 2y]
x2 = ax2/a +8a2 + 2ay
x2 = x2 + 8a2 + 2ay
2ay = -8a2
y = -4a

is that right then?</sup>

^{Just be careful when expanding several sets of brackets.
y=-4a should be the locus.}