parmetrics 2007 hsc question (1 Viewer)

[yugi]

Question:
The diagram shows a point P(2ap,ap^2) on the parabola x^2=4ay
The normal to the parabola at P intersects the parabola again at Q(2aq, aq^2)
The equation of PQ is x-py-2ap-ap^3=0 (DO NOT PROVE THIS)

i) Prove that p^2+pq+2=0
ii) If the chords OP and OQ are perpendicular, show that p^2 =2

 

[addikaye03]

Question:

The diagram shows a point P(2ap,ap^2) on the parabola x^2=4ay
The normal to the parabola at P intersects the parabola again at Q(2aq, aq^2)
The equation of PQ is x-py-2ap-ap^3=0 (DO NOT PROVE THIS)

i) Prove that p^2+pq+2=0
ii) If the chords OP and OQ are perpendicular, show that p^2 =2

Sub in (2aq, aq^2)

(2aq)-p(aq^2)-2ap-ap^3=0

2aq-apq^2-2ap-ap^3=0

2q+pq^2-2p-p^3=0

2(p-q)+p(q^2-p^2)

2(q-p)+p(q-p)(q+p)=0

(q-p)[2+p(p+q)]=0

2+pq+p^2=0

ii) OP and OQ are perpendicular [Mop x Moq=-1]

p/2 x q/2=-1

therefore pq=-4

sub into i) p^2-4+2=0 therefore p^2=2 as required

 

[alchetor]

HELLO YOOSHI

Anyway, here are your soln:

i)RTP: p^2 + pq + 2 = 0

EQ of parabola: x^2 = 4Ay
y= x^2/4a
dy/dx = x/2a
therefore, gradient at P is: p , and hence MNormal at P = -1/P

Also, the gradient of PQ is found by: q(p^2 - q^2) / 2a(p-q) [ gradient formula ]

Hence, by equating both forms of the gradient:
-1/p = (p^2 - q^2 ) / 2(p-q)
-1/p = (p-q)(p+q) / 2(p-q)
-2 = p(p+q)
p^2 + pq + 2 = 0
QED.

ii)Since, OP is perpendicular to OQ,
MOP x MOQ = -1
and
MOP = ap^2/2ap = p/2
MOQ = aq^2/2aq = q/2

therefore, pq/4 = -1
pq = -4
hence, sub pq=-4 into p^2 + pq + 2 =0
p^2 = 2


Lol sorry about all the words, i cbb finding symbols to represent them xD

edit: LOL dam the person above me just posted x_x'

 

[yugi]

bernard? :L

THANKS GUYS (Y)