Question 3448: 1 Eliminate, x,y between the following equations
x^2-y^2=px-qy, 4xy=qx+py , x^2+y^2=1
2. Eliminate, x,y,z between equations
y/z - z/y = a , z/x -x/z = b , x/y - y/x =c
Answer by khwang(438)   (Show Source):
You can put this solution on YOUR website! Hi, somehow this is not an easy problem about elimination.
This first one should use formulas in Trig., for otherwise, it is
almost impossible to get the cancellation of x,y & z.
1 Eliminate, x,y between the following equations
x^2-y^2=px-qy...(1), 4xy=qx+py...(2) , x^2+y^2=1...(3).
Sol: Since (3) is the unit circle x^2 + y^2=1,
x = cos t, y = sin t for some t.
Goto (1), we have cos^2 t- sin^2 t = (By double angle formula)
cos 2t = p cos t - q sin t = sqrt(p^2+q^2) [p cos t/sqrt(p^2+q^2) - qsin t/sqrt(p^2+q^2)]
= sqrt(p^2+q^2) cos (w + t)....(4) where cos w = p/sqrt(p^2+q^2)
Goto (2), 4 cos t sin t = q cos t + p sin t
= sqrt(p^2+q^2) [q cos t/sqrt(p^2+q^2) + p sin t/sqrt(p^2+q^2)]
So, 2 sin 2t = sqrt(p^2+q^2) sin (w + t) ...(5) [Note cos w = p/sqrt(p^2+q^2)]
(4)^2 + (5)^2 : cos^2 2t + 4 sin^2 2t = (p^2+q^2) [cos^2 (w + t) + sin^2 (w + t)]
Simplify both sides: 1 + 3 sin^2 2t = p^2 + q^2 [we can get the same relation by using (1)^2+(2)^2 ]
or sin^2 2t = (p^2 + q^2 - 1)/3 or 4 sin^2 t cos^2 t = (p^2 + q^2 - 1)/3
Since sin^2 2t = 4 (sin^2 t/ cos^2 t) cos^4 t = 4 tan^2 t /sec ^4 t = 4 tan^2 t /(1+tan^2 t )^2,
we have tan^2 t /(1+tan^2 t )^2 = (p^2 + q^2 - 1)/12 or
[(1+tan^2 t )^2/tan^2 t] = 12/(p^2 + q^2 - 1) or
So, (1+tan^2 t )/tan t = sqrt[12/(p^2 + q^2 - 1)] or
tan t + 1/tan t = sqrt[12/(p^2 + q^2 - 1)] ...(6)
Let r = sqrt[12/(p^2 + q^2 - 1)] = 2sqrt[3/(p^2 + q^2 - 1)]
Convert (6) to the quuadratic equation u^2 -ru + 1 = 0 , where u = tan t,
by quadratic formula u = tan t = [r + sqrt(r^2-4)]/2 [or (r - sqrt(r^2-4))/2]
Hence, tan t = [r+sqrt(r^2-4)]/2 = u ...(7)
Consider (5)/(4): 2 tan 2t = tan (w+t), so
4 tan t/(1 - tan^2 t) = (tan w + tan t) / (1 - tan w tan t)
Cancelling the denominator and simplify:
tan^3 t - 3 tan w tan ^2 t + 3 tan t - tan w = 0 ...(8)
Replace tan t by (7) in (8) , we obtain
[Note tan w = q/p, and so tan t = y/x , x, y disappear.]
[r+sqrt(r^2-4)]^3/8 - 3 q/p^2/4 + 3[r+sqrt(r^2-4)]/2 - q/p =0.
{Recall : r = 2sqrt[3/(p^2 + q^2 - 1)] ,r^2 = 12/(p^2 + q^2 - 1)}
{You can try to simplify it. But, it seems that there is no simple algebraic
form of for it. In other words, it is in complicated and ugly form.]]
2. Eliminate, x,y,z between equations
y/z - z/y = a , z/x -x/z = b , x/y - y/x =c.
Let y/z = u, z/x = w, x/y = wu.
So, u -1/u = a,
or u^2 - au -1 =0
By quadratic formula u = [a+ sqrt(a^2+4)]/2 = y/z...(1)
[if u=y/z >0, otherwise u= (a- sqrt(a^2+4))/2 ]
Similary, w = [b+ sqrt(b^2+4)]/2 = z/x...(2) [if w=z/x >0]
and x/y = [c + sqrt(c^2+4)]/2 ...(3) [if x/y >0]
Since y/z * z/x * x/y = 1, we obtain
[a + sqrt(a^2+4)] [b + sqrt(b^2+4)][c+ sqrt(c^2+4)]/8 = 1,
[a + sqrt(a^2+4)] [b + sqrt(b^2+4)][c+ sqrt(c^2+4)] = 8.
More precisely,
Case(i) y/z, z/x > 0
[a + sqrt(a^2+4)] [b + sqrt(b^2+4)][c+ sqrt(c^2+4)] = 8.
Case(ii) y/z > 0, z/x < 0
[a + sqrt(a^2+4)] [b - sqrt(b^2+4)][c- sqrt(c^2+4)] = 8.
Case(ii) y/z < 0, z/x > 0
[a - sqrt(a^2+4)] [b + sqrt(b^2+4)][c- sqrt(c^2+4)] = 8.
Case(iii) y/z < 0, z/x < 0
[a - sqrt(a^2+4)] [b - sqrt(b^2+4)][c+ sqrt(c^2+4)] = 8.
Try to read carefully about every step.
Good luck !
Kenny
|
|
|
