Parabola as a locus - questions (1 Viewer)

[Petinga]

could someone post step by step

1. A parabolic arch has a span of 32cm and a height of 10cm. Calculate the height of the focus above the horizontal level through the end points.

2. Sjow the following represent parabolas and find vertex, focus and equation of the directrix:
a) y^2 - 4x +2y - 3 = 0
b)y^2 + 8x -4y -8 = 0

3. Find the equation of the following parabolas and sketch each:
a) focus at (0,7), directrix the line y=-7
b) the line x=2 as axis, vertex (2,5) and crosses y-axis at y=9

4. Write equations in form (x-h)^2=4a(y-k) and hence state the vertex, focus, directrix and axis of the parabola:
a)y= x^2 - 4x +4
b)y= x^2 + 6x +6

 

[littleboy]

1. half of the span is 16cm being 10cm high. Therefore let that point be on the parabola P(16,10).
let the focus be S(0,a) and the directrix be y= -a

we know that the distance from the focus to any point on the parabola equals to its distance to the directrix.

draw this info and we know that:
(10-a)^2 + 16^2 = (10 + a)^2 (using pythagoras)

expand and simplify and u get: 40a = 256
a = 6.4
therefore focus is 6.4cm above horizontal level.

(not sure if all this is right, plz check)






2a. y^2 +2y+1 = 3+1+4x
(y+1)^2 = 4 +4x
(y+1)^2 = 4(x+1)
therefore vertex is: (-1,-1)
since it is in form y^2 = 4ax
a = 1
and hence focus : (0,-1)
directrix: x = -2

do the same for part b
i recommend drawing a diagram to find the focus and directrix

 

[littleboy]

4a. y = (x-2)(x-2) - factorise
y = (x-2)^2

This is in the form (x-h)^2 = 4a(y-k), where a = 1/4
vertex: (2,0)
focus: (2, 1/4)
directrix: y = -1/4
and the axis is the y axis

b) complete the square:
y = (x+3)^2 - 3
y+3 = (x+3)^2
this is then in the form (x-h)^2=4a(y-k)
where a = 1/4
therefore vertex: (-3,-3)
focus: (-3, -11/4)
directrix : y = -13/4

 

[Riviet]

i'll do the non-graphical bit of Q3.
Q3 a) We know the distance from the focus to vertex is equal to the distance from the directrix to the focus. Therfore y co-cordinate of vertex is the midpoint of the y co-ords of focus and directrix
=>y=(7-7)/2
=0
.: vertex is at origin (0,0)
.: a=7
.: equation is x²=28ay

b) Using the definition (x-h)²=4a(y-k)
vertex is (h,k)
sub in vertex (2,5)
=>(x-2)²=4a(y-5)
when x=0, y=9 -> sub these values in
=> 4=4a(9-5)
16a+4
.:a=1/4
.: equation is (x-2)²=y-5