What is the length of the longest possible laser beam he could send rounded to the nearest tenth of an inch?
1 answer:
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We know that
case a)
the equation of the vertical parabola write in vertex form is
y=a(x-h)²+k,
where (h, k) is the vertex.
Using our vertex, we have:
y=a(x-2)²-1
We know that the parabola goes through (5, 0),
so
we can use these coordinates to find the value of a:
0=a(5-2)²-1
0=a(3)²-1
0=9a-1
Add 1 to both sides:
0+1=9a-1+1
1=9a
Divide both sides by 9:
1/9 = 9a/9
1/9 = a
y=(1/9)(x-2)²-1
the answer is
a=1/9
case b)
the equation of the horizontal parabola write in vertex form is
x=a(y-k)²+h,
where (h, k) is the vertex.
Using our vertex, we have:
x=a(y+1)²+2,
We know that the parabola goes through (5, 0),
so
we can use these coordinates to find the value of a:
5=a(0+1)²+2
5=a+2
a=5-2
a=3
x=3(y+1)²+2
the answer is
a=3
see the attached figure
case a)
the equation of the vertical parabola write in vertex form is
y=a(x-h)²+k,
where (h, k) is the vertex.
Using our vertex, we have:
y=a(x-2)²-1
We know that the parabola goes through (5, 0),
so
we can use these coordinates to find the value of a:
0=a(5-2)²-1
0=a(3)²-1
0=9a-1
Add 1 to both sides:
0+1=9a-1+1
1=9a
Divide both sides by 9:
1/9 = 9a/9
1/9 = a
y=(1/9)(x-2)²-1
the answer is
a=1/9
case b)
the equation of the horizontal parabola write in vertex form is
x=a(y-k)²+h,
where (h, k) is the vertex.
Using our vertex, we have:
x=a(y+1)²+2,
We know that the parabola goes through (5, 0),
so
we can use these coordinates to find the value of a:
5=a(0+1)²+2
5=a+2
a=5-2
a=3
x=3(y+1)²+2
the answer is
a=3
see the attached figure