SOLUTION: Dear Sir, Please I beg you to help me for this one question. Find the equation of the parabola with vertex at (0, 0) and focus at (1/8,0). Sketch and determine the parts of t

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Dear Sir, Please I beg you to help me for this one question. Find the equation of the parabola with vertex at (0, 0) and focus at (1/8,0). Sketch and determine the parts of t      Log On


   

Question 1181123: Dear Sir,
Please I beg you to help me for this one question.
Find the equation of the parabola with vertex at (0, 0) and focus at (1/8,0). Sketch and determine the parts of the parabola.
Equation of the Parabola
Graph of the Parabola
Parts of the Parabola
1. Vertex
2. Focus
3. Directrix
4. Axis of Symmetry
5. 𝑬𝟏, 𝑬𝟐
6. Length of 𝐸1, 𝐸2
7. Graph
Thank you so much for your help.

God bless you.
Lorna

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
given:
vertex at (0, 0)=>h=0, k=0
and focus at (1%2F8,0).
Since the y+values are the same, use the equation of a parabola that opens left or right.
%28y-k%29%5E2=4p%28x-h%29
%28y-0%29%5E2=4p%28x-0%29
y%5E2=4p%2Ax-> disance from the focus to the vertex is p and in your case p=1%2F8

y%5E2=4%281%2F8%29%2Ax
your equation is:
y%5E2=%281%2F2%29%2Ax

E1, E2 lie on intersection of a line x=1%2F8 with parabola
y%5E2=%281%2F2%29%2A%281%2F8%29
y%5E2=1%2F16
y=sqrt%281%2F16%29
y=1%2F4 or y=-1%2F4
E1 is at (1%2F8, 1%2F4)
E2 is at (1%2F8, -1%2F4)
1. Vertex: (0, 0)
2. Focus:(1%2F8, 0)
3. Directrix: x+=+-1%2F8
4. Axis of Symmetry: y axis
5. E1, E2
E1 is at (1%2F8, 1%2F4)
E2 is at (1%2F8, -1%2F4)
6. Length of E1, E2: 2
Solved by pluggable solver: Distance Between 2 points
The distance formula is sqrt%28%28%28x%5B2%5D-x%5B1%5D%29%5E2%29%2B%28%28y%5B2%5D-y%5B1%5D%29%5E2%29%29. Plug in the numbers,
sqrt%28%28%280.125-%280.125%29%29%5E2%29%2B%28%28-0.25-%280.25%29%29%5E2%29%29
sqrt%280%5E2%2B-0.5%5E2%29 The distance is 0.5.



7. Graph