SOLUTION: show that the equation of the parabola whose vertex and focus are on the x-axis at distances a and a' from the origin respectively is y^2=4(a'-a)(x-a).

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: show that the equation of the parabola whose vertex and focus are on the x-axis at distances a and a' from the origin respectively is y^2=4(a'-a)(x-a).       Log On


   

Question 827582: show that the equation of the parabola whose vertex and focus are on the x-axis at distances a and a' from the origin respectively is y^2=4(a'-a)(x-a).
Answer by lwsshak3(11628) About Me  (Show Source):
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show that the equation of the parabola whose vertex and focus are on the x-axis at distances a and a' from the origin respectively is y^2=4(a'-a)(x-a).
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given parabola opens rightward with axis of symmetry: y=0
its standard form of equation: (y-k)^2=4p(x-h), (h,k)=(x,y) coordinates of the vertex
p=(a'-a)
4p=4(a'-a)
x-coordinate of vertex=a
y-coordinate of vertex=0
equation:y^2=4(a'-a)(x-a)