Question

show that for all values of m, the line y=mx-3m^2 touches the parabola x^2=12y

Answer 1

find the point of intersection by substituting the linear equation into the parabola... and solve for x...

Answer 2

the number of solutions will determine if it touches in cuts twice

Answer 3

oops or doesn't intersect at all

Answer 4

u mean sub y into the x square = 12 y ?

Answer 5

this is what I mean

\[x^2 = 12(mx - 3m^2)\]

Answer 6

ok coni=tinue

Answer 7

then do what ?

Answer 8

show not draw lol , its a calculator free paper dude

Answer 9

you have to show that the line then is the tangent line of parabola

slope of tangent line = dy/dx

dy/dx = x/6

tangent line:
Y = (x1/6)(X-x1) + y1 for some point (x1,y1)
Y = (x1/6)X + (y1 -x1^2/6)

and y1 = x1^2/12

--> Y = (x1/6)X -x1^2/12 = mX -3m^2

m = x/6

for every x value there is a "m" value, thus the line is tangent to parabola for all "m"
sorry if this is hard to read

Answer 10

so then it becomes

\[x^2 = 12mx - 36m^2\]

or

\[x^2 - 12mx + 36m^2 = 0\]

so now solve for x by factoring

Answer 11

ok wait for a while

Answer 12

after I got x then what again ?

Answer 13

ans when you factor it you get

\[(x - 6m)^2 = 0\]

solve for x, which seems to show the 2 curves have 1 common point of contact...

Answer 14

so you have proven the case...

Answer 15

ok