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2. The cable of a suspension bridge hangs in the shape of a parabola.
The tower supporting the cable are 400 ft apart and 150 ft high.
If the cable, at its lowest, is 30 ft above the bridge at its midpoint,
how high is the cable 50 ft away(horizontally) from either tower?
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Place the origin of the coordinate system half way between the towers,
at the level of the bridge.
Using vertex form, write equation of a parabola
y = ax^2 + 30.
Find the unknown coefficient "a" from the condition
a*200^2 + 30 = 150 ft,
which says that the cable is suspended at the top of a tower (for each cable and each tower).
From this equation, find "a"
40000a = 150 - 30 = 120
a = = = 0.003.
So, the parabola equation is
y = 0.003*x^2 + 30.
They want you find "y" at x = 200-50 = 150 ft.
Substitute x= 150 into equation of the parabola and get
y = 0.003*150^2+30 = 97.5 ft.
ANSWER. 50 ft away from either tower, the cable is 97.5 ft over the bridge.
Solved.
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Ikleyn did the parabola problem for you. I'll do the circle problem.
1.Find the equation of the circle that passes through the points (0,0), (5,0),
and (3,3). Use both geometric and algebraic solution for comparison.OK, I'll try to show the straight-edge and compass geometric method to draw
a circle through three points.
The geometric method is to find the center by:
1. drawing two chords,
2. drawing their perpendicular bisectors (in green):
3. the center is where those two perpendicular bisectors of 2 chords intersect.
(it looks like the point , but we aren't sure
of it being exactly that with the geometric method.)
4. Put the foot of the compass on the point where the perpendicular bisectors
of the two chords intersect. The open your compass to one of the points on
the circle. Then swing it around and draw a circle.Â
That's the geometric way to draw a circle through three points.
But the algebraic way is to begin with the general equation
or the standard equation for a circle:'
It might be easier to substitute the three points in the standard equation:
Subtract the 1st equation from the other two equations:
Solve the 2nd equation for h
Substitute 5/2 for h in the 3rd equation
Multiply the 3rd equation by 2
Substitute h=5/2, k=1/2 in the 1st equation:
So the standard form is
So now we see that the center = (h,k) =
and the radius is
So we graph the circle
Other times we might just want the general equation
Substitute the three points in:
Substitute c=0 in the other two equations:
Divide the 2nd equation by 5 and the third by 3
Solve the 2nd equation for 'a'
Substitute a = -5 in the third equation:
Solve the 3rd equation for b
Substitute in the general solution:
But the trouble with the general equation of a circle is that we can't tell
the center and radius from it unless we complete the square and get it into
the standard form.
The general form and the standard form
are equivalent.
Edwin