Show that for all values of m, the line y=mx-3m^2 touches the parabola x^2=12y
please help me!
thanks!
2007-11-25 22:10:36 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
X^2=12(mx-3m^2)
x^2=12mx-36m^2
x^2-12mx+36m^2=0
x=6m+sqrt(144m^2-144m^2)/2
I get it is true only when x=6m y=3m^2;
Therefore y=x^2/12; thus x^2=12y. Yeah! I got it.
2007-11-25 22:23:04 · answer #1 · answered by ramesh_1960 3 · 0⤊ 0⤋
Take the equation for the line:
y = mx - 3m²
Plug this into the equation for the parabola:
x² = 12y
x² = 12(mx - 3m²)
x² = 12mx - 36m²
Get everything on one side:
x² - 12mx + 36m² = 0
Factor:
(x - 6m)² = 0
Now imagine any value of m. No matter what there will be a value of x that will satisfy this equation. (Specifically x = 6m).
So regardless of the value of m, the two functions will intersect at some point.
2007-11-25 22:32:24 · answer #2 · answered by Puzzling 7 · 0⤊ 0⤋
substibute 1st equ into 2nd equ
x^2=12(mx-3m^2)
x^2=12mx-36m^2
36m^2-12m+x^2=0
(6m-x)^2=0
m=+-x/6
2007-11-25 22:19:04 · answer #3 · answered by someone else 7 · 0⤊ 0⤋