SOLUTION: Find the standard form of the equation of the parabola with the given characteristic(s) and vertex at the origin. Horizontal axis and passes through the point (−2, 5)

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Find the standard form of the equation of the parabola with the given characteristic(s) and vertex at the origin. Horizontal axis and passes through the point (−2, 5)      Log On


   

Question 1136722: Find the standard form of the equation of the parabola with the given characteristic(s) and vertex at the origin.
Horizontal axis and passes through the point (−2, 5)
Found 2 solutions by josgarithmetic, greenestamps:
Answer by josgarithmetic(39618) About Me  (Show Source):
You can put this solution on YOUR website!
x-h=a%28y-k%29%5E2, horizontal symmetry axis, vertex (h,k)

x=ay%5E2, vertex is (0,0).

a=x%2Fy%5E2

a=-2%2F25, using the given point on the parabola.

highlight%28x=-%282%2F25%29y%5E2%29

Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


y=ax^2 has a vertical axis of symmetry; a horizontal axis of symmetry with vertex at the origin means the equation is x = ay^2.

Find the value of a by substituting the coordinates of the given point in the equation.

-2+=+a%285%5E2%29
-2+=+25a
a+=+-2%2F25

The equation is

x+=+%28-2%2F25%29y%5E2

A graph, showing the parabola passing through (-2,5)...