Eliminate the parameter t to find a Cartesian equation for:
x=t^2
y=3+4t
x=Ay^2+By+C
Where A = ?
B = ?
C= ?
2007-10-12 17:23:35 · 4 answers · asked by snowwind_7 2 in Science & Mathematics ➔ Mathematics
thank you!
2007-10-12 17:38:16 · update #1
x=t^2
x=[(y-3)/4]^2=(y-3)^2/16
y=3+4t
t=(y-3)/4
x=Ay^2+By+C
(y-3)^2/16=Ay^2+By+C
y^2-6y+9=16Ay^2+16By+16C
coefficient of y^2
1=16A
A=1/16
coefficient of y
-6=16B
B=-6/16=-3/6
coefficient off K
9=16C
C=9/16
2007-10-12 17:33:01 · answer #1 · answered by ptolemy862000 4 · 0⤊ 0⤋
from y = 3 + 4t t = (y-3)/4
x = ((y-3)/4) ^ 2
= (y-3)^2 / 16
= (y^2 - 6y + 9)/16
= 1/16 y^2 - 3/8 y + 9/16
A = 1/16, B= -3/8, C=9/16
2007-10-12 17:34:00 · answer #2 · answered by norman 7 · 0⤊ 0⤋
[x=t^2
y=3+4t
x=Ay^2+By+C
//given]
y-3=4t
(y-3)/4=t
subsitute in
x= ((y-3)/4)^2
x=(y-3)^2*1/16
x = (y^2 -6y +9)/16
A=1/16
B= -3/8
C= 9/16
2007-10-17 04:41:34 · answer #3 · answered by Matt B 3 · 0⤊ 0⤋
t = (y - 3)/4
x = [(y - 3)/4]^2
x = (y^2 - 6y + 9)/16
A = 1/16
B = - 3/8
C = 9/16
2007-10-12 18:17:21 · answer #4 · answered by Helmut 7 · 0⤊ 0⤋