James Bennett

Posted on Friday, 21 March, 2003 - 06:23 pm:

Please help - I do not know how to answer the following 2 questions; I have tried different ways but do not get the right answers! Please explain how to work them out and give relevant answers [at least the answers to 1 of the q's since I think they are quite similar methods]
1. Find the gradients of the curves y^2=2x and x^2=2y at their points of intersection. Hence find the angles at which the curves intersect [N.B. tan theta= [m1 - m2]/[1 + m1m2] ]
2. Find the points of intersection of xy=1 and x^2-2y^2+1=0 and determine the angles at which the curves intersect.

*** '^' = to power of. ***
Thanks!

Kerwin Hui

Posted on Friday, 21 March, 2003 - 06:55 pm:

1. The points of intersections are (0,0) and (2,2). At (0,0), it is obvious that the curves intersect orthogonally (tangents are the coordinate axes). At (2,2), we get m₁=2, m₂=1/2, giving tanq = (3/2)/2=3/4. so q = 36.9 degrees.

Kerwin

James Bennett

Posted on Saturday, 22 March, 2003 - 07:47 pm:

I don't mean to be silly but how are we finding the points of intersection? I put the 2 equations equal to each other but can't seem to solve them!

James Bennett

Posted on Saturday, 22 March, 2003 - 07:55 pm:

By solving them implicitly I still can't seem to do this!

Alice Thompson

Posted on Saturday, 22 March, 2003 - 08:25 pm:

You can find points of intersection of both using substitution.

For the second one the points of intersection are (1,1) and (-1,-1). The gradients of the curves are -y/x and x/2y, so the angle between the curves is the same at both points. Using this, tanq = -3, so q = 71.6 degrees.

James Bennett

Posted on Saturday, 22 March, 2003 - 10:10 pm:

Could somebody go through the substitution procedure step for finding curve points of intersection for the 2 q's above pls?

Kerwin Hui

Posted on Saturday, 22 March, 2003 - 10:30 pm:

1. 8x=4y² =(2y)² =(x² )² =x⁴ , so x=0,2, and it is easy to find the corresponding y.

2. xy=1 gives y=1/x (x =/= 0). Substitute in the other equation gives x² -2x^-2 +1=0, so x² =-2 or 1. Rejecting -2 gives x=±1, and again it is easy to find the corresponding y.

Kerwin

James Bennett

Posted on Saturday, 22 March, 2003 - 10:33 pm:

Thanks!

Alice Thompson

Posted on Saturday, 22 March, 2003 - 10:38 pm:

For the first one, y^2=2x and x^2=2y, x=y^2/2, so at the intersection, y^4/4=2y, then y(y^3-8)=0.

Putting this back in the original equation gives (0,0) and (2,2)

For the second, x^2-2y^2+1=0 and xy=1. Using y=1/x you get

x^2-2x^-2+1=0, x must be positive at the point of intersection, so

x^4+x^2-2=0
(x^2+2)(x+1)(x-1)=0

Putting this back in the equation gives (1,1) and (-1,-1)

James Bennett

Posted on Sunday, 23 March, 2003 - 12:57 pm:

Thanks everyone for your input!