Help with Circle Geometry and Parametrics? (1 Viewer)

[sk8ie_boi]

Some help please??

Thanks

 

[FinalFantasy]

a) after u draw it, u should recognise dat tangent to a circle makes 90 degrees w\ it's radius. and from T to where it meets the circle, it's 6.
so line from T to center of the circle is sqrt(6²+2.5²)
so radius of da circle center T is sqrt(6²+2.5²)-2.5 cm

b)given eq. of normal at P(2ap, ap²): x+py=ap³+2ap---->(1)
x²=4ay
y=x²\4a
sub. y into (1):
x+p(x²\4a)=ap³+2ap
4ax+px²=4a²p³+8a²p
px²+4ax-(4a²p³+8a²p)=0

x=[-4a+-sqrt(16a²+4p(4a²p³+8a²p) )]\2p
=[-4a+-sqrt(16a²+16a²p^4+32a²p²)]\2p
=[-4a+-4sqrt(a²+a²p^4+2a²p²)]\2p
=[-2a+-2asqrt(1+p^4+2p²)]\p
=[-2a+-2asqrt((p²+1)²)]\p
=[-2a+-2a(p²+1)]\p
=-2a(1+-(p²+1))\p=-2a(2+p²)\p just taking one of dem
y=x²\4a=4a²(2+p²)²\4ap²=a(2+p²)²\p²

x=2ap
x=-2a(2+p²)\p since p is parameter

y=ap²
.: y=a[-(2+p²)²\p²]=a(2+p²)\p²
.: this normal meets the parabola again at the point w\ parameter -(2+p²)\p

 

[ngai]

b) normal is x+py=ap³+2ap, parabola is y=x²/4a

these two meet when:
x+px²/4a=ap³+2ap

when x=2a(-2-p²)/p,
LHS = ... (sub it in)
= ... (expand)
= ap³+2ap
= RHS

so the pt corresponding to parameter -(2+p²)/p is the other intersection

 

[sk8ie_boi]

x=2ap
x=-2a(2+p²)\p since p is parameter

y=ap²
.: y=a[-(2+p²)²\p²]=a(2+p²)\p²
.: this normal meets the parabola again at the point w\ parameter -(2+p²)\p

Oh thanks man, umm .. I get everything else .. I just don't get this bit...

 

[FinalFantasy]

wow, ngai's solution is much better and clearer, i'm not even sure if i can do wat i did.
just look at ngai's one lol

 

[Trebla]

Hey ngai, are you by any chance Andrew Tin-Pui Ngai from James Ruse Agricultural High School, who topped Mathematics Extension 2 in last year's HSC?

 

[FinalFantasy]

this ngai is indeed the master! the number 1 maths extension 2 student of 2004 lol

 

[sk8ie_boi]

Yepz .. I got it now .. thanks yous .. hehe