dnt stop runnin
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how do u do question 8)d..........i think i was completely of track for it
Question 8)d............who did it and how?? (1 Viewer)
- Thread starter dnt stop runnin
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ok, what i did was prove that y= mx-3m^2, has real roots, then prove its a parabola cuz it got 1 stat. pt. , thus its domain is all real x, therfore ti must touch the parabola, ....
i dunno thats what i thought
i dunno thats what i thought
(i) you get y=(x^2)/12 from the second equation and sub it into the first one:
thus, x^2 - 12mx - 36m^2 = 0
if the line touches the parabola (ie it is a tangent to it), then the discriminant = 0
discriminant = b^2 - 4ac
=144m^2 - 4(1)(-36m^2)
=0
therefore, the line y = mx - 3m^2 touches the parabola x^2=12y
(ii) if the line y = mx - 3m^2 passes through (5,2) then x=5 and y=2 must satisfy the equation
so 2 = 5m - 3m^2
3m^2 - 5m - 2 = 0
(3m+1)(m-2) = 0
therefore m = -1/3 or 2
the line y=mx-3m^2 passes through (5,2) when m=-1/3 or m=2
(iii) sub the values of m in and you get,
y=2x-12 and
y=-x/3 - 1/3
thus, x^2 - 12mx - 36m^2 = 0
if the line touches the parabola (ie it is a tangent to it), then the discriminant = 0
discriminant = b^2 - 4ac
=144m^2 - 4(1)(-36m^2)
=0
therefore, the line y = mx - 3m^2 touches the parabola x^2=12y
(ii) if the line y = mx - 3m^2 passes through (5,2) then x=5 and y=2 must satisfy the equation
so 2 = 5m - 3m^2
3m^2 - 5m - 2 = 0
(3m+1)(m-2) = 0
therefore m = -1/3 or 2
the line y=mx-3m^2 passes through (5,2) when m=-1/3 or m=2
(iii) sub the values of m in and you get,
y=2x-12 and
y=-x/3 - 1/3