Posted by mike on Tuesday, October 5, 2010 at 10:58pm.
The point (0,1) lies in the "interior" of the parabola
y = 3x^2.
There is no tangent to the curve that will pass through (0,1).
check your typing.
there are two tangents lines to the curve f(x) = 3x^2 that pass through the point p =0,-1 find the x coordinates of the point where the tangents line intersect the curve, please show working.
ahhh, now it makes sense.
let the point of contact be (a,b)
slope of tangent by the grade 9 way = (b+1)/a
slope of tangent by Calculus is
dy/dx = 6x
so at the point (a,b), slope = 6a
then 6a = (b+1)/a
6a^2 = b+1
but since (a,b) lies on the curve, b = 3a^2
so
6a^2 = 3a^2 + 1
a^2 = 1/3
a = ± 1/√3
so the x coordinates of the two tangents are 1/√3 and -1/√3
find the equation of the tangent line of the curve y=root x divideed by x+1 at the point p = 2, root 2 divided by 3. show working please.
find the equation of the tangent line to the curve y = 2√2/x+1 at the point p=1,2/3.please show working. i would appreciate if you do it before 12:05pm today
use implcit differenciation to find an eqaution of both the tangent line to the ellipse:
2x^2 + 4y^2 = 36
that passes through the points: 14,3
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