Given the parabola $$C:x^2+y-1=0$$ and the straight line $$x-2y-1=0$$ we call A, B to the intersection points. When the point P moves over the arc AB (included in C), find the maximum of the area of the triangle PAB and the value of P in that moment.
#Analytic geometry and area
eternal shell
finite wadiBOT
Given the parabola $$C:x^2+y-1=0$$ and the straight line $$x-2y-1=0$$ we call A,...
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rose caveBOT
roi
versed path
Heres what ibe got
What's the area of a triangle with given vertices (x1, y1), (x2, y2) and (x3, y3)?
eternal shell
its with determinants or a vector cross product right?
versed path
its with determinants or a vector cross product right?
Yeah.
shell grove
you can also use the bh/2 formula
base is constant, height above it will be maximized when point is moving parallel to the base
versed path
you can also use the bh/2 formula
base is constant, height above it will be maxi...
Well, not necessarily maximized, but that should be a critical point.
(in general, I mean)
shell grove
sure, yeah
eternal shell
Yeah.
i've got the area with the alpha variable, but how do i know its value when it is maximum now?
btw, the solution gets the area this way, but i dont really understand how they do it
also, if i arrange the order of the points differently in the determinant, i get different values, how do i know which one is correct?
something like this, in one i get +5/4 and in the other one i get -5/4
why?
versed path
also, if i arrange the order of the points differently in the determinant, i get...
Note that changing the order of rows in a determinant changes its sign. Absolute value is thus also needed.
But since optimizing absolute value isn't easy, you can optimize S^2 instead of S.
versed path
So, in other words, we have:
r(A) = {1, 0}
r(B) = {-3/2, -5/4}
r(P) = {t, 1 - t^2}
So:
AB = {-5/2, -5/4} = -(5/4){2, 1}
AP = {t - 1, 1 - t^2} = (1 - t){-1, 1 + t}
S^2 = (1/4)(5/4)^2 (1 - t)^2 (2(1 + t) + 1)^2 = (25/64)(t - 1)^2 (2t + 3)^2
To make this more simple, we can use logarithmic differentiation.
ln(S^2) = ln(25/64) + 2ln(|t - 1|) + 2ln(|2t + 3|)
d(ln(S^2))/dt = 2/(t - 1) + 4/(2t + 3)
And the rest is easy.
eternal shell
So, in other words, we have:
r(A) = {1, 0}
r(B) = {-3/2, -5/4}
r(P) = {t, 1 - t^...
Thanks man
lofty lionBOT
@eternal shell has given 1 rep to @versed path