# Conics rectangle hyperbola (FP2)Watch

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#1
The hyperbola C has an equation of xy=c^2, x=ct, y=c/t

An equation of the normal to C at the point where t=p is py+c(p^4)=(p^3)x+c

How to verify that this normal meets C again at the point at which t=q, where q(p^3)+1=0 ???

I am not certain about what it really asks.. It wants me to make use of q(p^3)+1=0?? or wt?????
0
13 years ago
#2
It wants you to show that the normal intersects the hyperbola at a point other than t=p.

One way to do this is to let y=c^2/x and plug this into the equation of the normal:
p(c^2/x) + cp^4 = p^3x + c
=> pc^2 + (cp^4)x = (p^3)x^2 + cx
=> (p^3)x^2 + c(1 - p^4)x - pc^2 = 0

We know that this quadratic has x=cp as a root (since the normal intersects the curve at p). So if cq is another root, then:
cq * cp = (-pc^2)/p^3
=> p^3 q = -1
=> p^3 q + 1 = 0, as required.
0
13 years ago
#3
(Original post by im37)
The hyperbola C has an equation of xy=c^2, x=ct, y=c/t

An equation of the normal to C at the point where t=p is py+c(p^4)=(p^3)x+c

How to verify that this normal meets C again at the point at which t=q, where q(p^3)+1=0 ???

I am not certain about what it really asks.. It wants me to make use of q(p^3)+1=0?? or wt?????
You just need to find the other point at which the normal meets the hyperbola C.

So, rerrange xy = c2 to get x = c2/y (or you could rearrange for y just as easily) and sub into the equation of the normal.

py + cp4=p3x + c
=> py + cp4=p3(c2/y) + c

Re arrange for a quadratic in y:

py2 + (cp4 - c)y + (-c2p3)

Solve using the quadratic formula to get

y = c/p or y = -cp3

Now, y = c/t and t=q, so y=c/q

c/q = c/p will give p =q

and

c/q = -cp3
=>1/q = -p3
=>qp3 + 1 = 0
0
13 years ago
#4
(Original post by im37)
The hyperbola C has an equation of xy=c^2, x=ct, y=c/t

An equation of the normal to C at the point where t=p is py+c(p^4)=(p^3)x+c

How to verify that this normal meets C again at the point at which t=q, where q(p^3)+1=0 ???

I am not certain about what it really asks.. It wants me to make use of q(p^3)+1=0?? or wt?????
xy = c^2
x = ct = cp at t = p
y = c/t = c/p at t = p

dy/dx = -ct^-2/c = -1/t^2 = -1/p^2 at t = p

y - c/p = -1/(-1/p^2)(x-ct)
=> y - c/p = p^2(x-cp)
=> py - c = p^3x - cp^4
=> py + cp^4 = p^3x + c

Sub in values of x = ct and y = c/t to see where normal intersects curve.

p(c/t) + cp^4 = p^3(ct) + c
=> pc/t + cp^4 = cp^3t + c

Divide by c
=> p/t + p^4 = p^3t + 1
=> p + p^4t = p^3t^2 + t
=> p^3t^2 + (1-p^4)t - p = 0

t = [p^4-1 +/- rt((1-p^4)^2-4(p^3)(-p))]/2p^3
=> t = [p^4-1 +/- rt(1+p^4)^2)]/2p^3
=> t = (p^4-1 +/- 1+p^4)/2p^3
=> t = (2p^4)/2p^3 = p (point known) OR t = -2/2p^3 (at t = q)

=> at t = q
2p^3q = -2
=> p^3q = -1
=> p^3q + 1 = 0

A long method but it works!
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