Just part c please thanks
1 answer:
Unfortunately the portion of the graph shown does not show the locations of points A and B, nor does it show sufficient parts of the curve.
Fortunately, to do part C, it is not necessary to use the graph of the function, nor the locations of points A & B, since we know that t=-1 at A.
The solution is as follows:
We know that the tangent line has the equation
l : 2x -5y -9 = 0 ...................(1)
and that the parametric equation is
x = t^3 -8t; y = t^2 ..............(2)
We can obtain the intersections of the tangent line (1) with the original curve (2) by substituting (2) in (1), giving
f(t) = 2*(t^3-8*t) -5*(t^2)-9 = 0
or on simplification,
f(t) = 2t^3 -5t^2 -16t -9 = 0 ....................(3) parametric equation of intersection points
Solution of which gives t=9/2 or t=-1.
Since t=-1 is the given tangent point, we conclude that t=9/2 is point B.
Translating into rectangular coordinates for t=9/2, we get
x=t^3-8t=(9/2)^3-8(9/2)= 441/8 = 55.125
y=t^2=(9/2)^2= 81/4 = 20.25
Therefore the coordinates of B:
t=9/2, or (55.125, 20.25)
Attached is the graphics of f(t), showing that t=-1 is a tangent line, and that the solutions are t=-1, or t=9/2.
Fortunately, to do part C, it is not necessary to use the graph of the function, nor the locations of points A & B, since we know that t=-1 at A.
The solution is as follows:
We know that the tangent line has the equation
l : 2x -5y -9 = 0 ...................(1)
and that the parametric equation is
x = t^3 -8t; y = t^2 ..............(2)
We can obtain the intersections of the tangent line (1) with the original curve (2) by substituting (2) in (1), giving
f(t) = 2*(t^3-8*t) -5*(t^2)-9 = 0
or on simplification,
f(t) = 2t^3 -5t^2 -16t -9 = 0 ....................(3) parametric equation of intersection points
Solution of which gives t=9/2 or t=-1.
Since t=-1 is the given tangent point, we conclude that t=9/2 is point B.
Translating into rectangular coordinates for t=9/2, we get
x=t^3-8t=(9/2)^3-8(9/2)= 441/8 = 55.125
y=t^2=(9/2)^2= 81/4 = 20.25
Therefore the coordinates of B:
t=9/2, or (55.125, 20.25)
Attached is the graphics of f(t), showing that t=-1 is a tangent line, and that the solutions are t=-1, or t=9/2.
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so.... now once we know what C is
you can if you want, do a search in google for "inscribed quadrilateral conjecture", I can do a quick proof if you need one
but in short, for a quadrilateral inscribed in a circle, each pair of opposites angles are "supplementary angles", namely they add up to 180°
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