SOLUTION: Please show me how to solve this.... 1.)How high is a parabolic arc pf span 24m and height 18m at distance 8 from the center of it span? 2.)A chord passing through th

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Please show me how to solve this.... 1.)How high is a parabolic arc pf span 24m and height 18m at distance 8 from the center of it span? 2.)A chord passing through th      Log On


   

Question 576205: Please show me how to solve this....
1.)How high is a parabolic arc pf span 24m and height 18m at distance 8 from the center of it span?




2.)A chord passing through the focus of the parabola x^2=16y has one end at pt. (12,9). Where is the end of the other chord?
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
1.) Let's make x the horizontal distance (in m) from the center of the span (which would be right below the vertex of the parabola, the top of the arch).
Let's make y, the height, in meters.
The top of the arch would be the (0,18), the vertex of the parabola.
For all x between -12 and 12,
y=ax%5E2%2B18
That is the equation of the parabola in vertex form (because the vertex is (0,18).
We just need to find a, knowing that for x=12, and x=-12, y=0.
0=a%2A12%5E2%2B18 --> 0=144a%2B18 --> 144a=-18 --> a=-18%2F144 --> a=-1%2F8
So y=-x%5E2%2F8%2B18
I am going to assume that is the problem means to ask how high is the arc above a point on the ground that is 8 m from the center of the span.
At 8 meters from the center of the span, on the ground, x=8.
At that point,
y=-8%5E2%2F8%2B18=-8%2B18=10
2.) x%5E2=16y
The axis of symmetry is x=0, the y axis, because y has the same value for x and -x.
The vertex of the parabola is the point (0,0) on thay axis of symmetry.
The focus of the parabola will be a point (0,c), and the directrix will be the line x=-c.
The point of the parabola with y=c, is at a (vertically measured) distance 2c from horizontal directrix x=-c. By the definition of parabola, the (horizontally measured) distance from that point to focus (0,c) is the same 2c, so for x=2c, y=c, and substituting into the equation for thee parabola, we get
%282c%29%5E2=16c --> 4c%5E2=16c --> c=4
So the focus is at (0,4).
The chord that passes through the focus and through (12,9) has a slope of
slope=%289-4%29%2F%2812-0%29=5%2F12, so its equation is
y=5x%2F12%2B4 since its y-intercept is the focus (0,4).
The intersection points of that line and the parabola x%5E2=16y will be solutions of
x%5E2=16%285x%2F12%2B4%29 --> x%5E2=20x%2F3%2B64 --> (((3x^2=20x+192}}} --> (((3x^2-20x-192=0}}}
The quadratic formula says that

So the solutions are:
x=72%2F6=12 (for the point (12,9) given), and for the point we need
x=-32%2F6 --> highlight%28x=-16%2F3%29, which corresponds to
y=%285%2F12%29%28-16%2F3%29%2B4=-20%2F9%2B4=-20%2F9%2B36%2F9 --> highlight%28y=16%2F9%29
The other end of the chord is at (-16/3,16/9) (unless I made a mistake in the calculations, of course).