How to Solve for h When Choosing a Value of a in a Parabola Equation?

[aspirare21]

Homework Statement

Parabolas with vertex on the x-axis,with axis parallel to the y-axis,and with distance from focus to vertex fixed as "a".

the question is pick your own value of "a". Then for "a" value pick value of "h". What does it mean? I'm confuse.. T__T

Homework Equations

(x-h)² = ±4a(y-k)
(x-h)² = ±4ay

k=0

The Attempt at a Solution

(x-h)² =4ay
2(x-h) =4ay'

answer is (x-h) 2ay'

square both sides
(x-h)² = 4a²(y')²

substitute
4a²(y')² = 4ay
a(y')² = y

I don't know what will happen if i choose a value of "a" then it will generate a value of "h".. T__T

 

[ideasrule]

Is that all the question says? Is there another part to the question? If not, OK, a=2 and h=3. Any value of a and h except a=0 will satisfy the conditions given in the question.

 

[aspirare21]

I forgot.. the answer should be in differential equation. show that when you pick "a" you should derive "h". Kindly confusing to me..