SOLUTION: Find the equation of the parabola with vertex (-2,-1), focal length 2 and axis parallel to the y-axis

Question 1122881: Find the equation of the parabola with vertex (-2,-1), focal length 2 and axis parallel to the y-axis
Answer by Alex.33(110) (Show Source): You can put this solution on YOUR website!
We can first consider the parabola with focal length 2 and vertex (0,0).
It's easy to see(as the formula of parabola I learned in the case that vertex is the origin), that the equation is x^2=8y, or y=x^2/8.
If you are curious about the formula visit Wikipedia. https://en.wikipedia.org/wiki/Parabola#Definition_as_a_locus_of_points

Now let's move this parabola to get its vertex to (-2,-1). To move any point (x,y) 2 units left, 1 units down, to its new home (x', y').

So we have x'=x-2, y'=y-1. Rewrite: x=x'+2, y=y'+1.

So now how do we figure out the equation of our new formula? Well, the only way to take is to use our previous relationship of x and y(i.e., the previous formula), and the relationship of (x,y) and (x',y'), to get the relationship(equation) between x' and y', describing the new parabola, since(x', y'), rather than (x,y), is a spot on the new parabola.

Now you may have seen why I rewrited the equations of (x,y) and (x', y').
Substitute the rewrited stuff into y=x^2/8. We get (y'+1)=(x'+2)^2/8
Simplify it to the form you like. And remove the silly apostrophes. Boom! Done.

This is the way I understand(and many people understand) what we are actually doing when we transform graphs. We are, at the end of the day, figuring out one relationship based on two relationships we previously have.

If you do think this is beneficial for you and many other people and want me to write a lesson on this, just send me a message. Thanks.

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