SOLUTION: A.) Write the equation of the parabola in general form completely draw thr graph. A. V(4,2) , F(2,2) B. V(5,-1) , F(3,-1) C. V(4,-4) , y+5=0 B.) Express each in standard form

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: A.) Write the equation of the parabola in general form completely draw thr graph. A. V(4,2) , F(2,2) B. V(5,-1) , F(3,-1) C. V(4,-4) , y+5=0 B.) Express each in standard form      Log On


   



Question 1039314: A.) Write the equation of the parabola in general form completely draw thr graph.
A. V(4,2) , F(2,2)
B. V(5,-1) , F(3,-1)
C. V(4,-4) , y+5=0
B.) Express each in standard form completely draw the graph
D. 4x²+24x-y+3=0
E. y²+4x+6y+17=0

Answer by josgarithmetic(39618) About Me  (Show Source):
You can put this solution on YOUR website!
Only doing one of those for you.

C. V(4,-4), y+5=0


This might be vertex and directrix. Solving for y gives y=-5 and no restriction on x for the directrix. This can also be stated as a variable point (x,-5), the directrix.

The focus is on the other side of the vertex. From -4 to -5 is a distance of 1. The focus is therefore 1 unit away from the vertex and this focus must be (4,-3).

Use the definition of a parabola. Find the equation for Distance from (x,y) to (x,-5) is equal to Distance from (x,y) to (4,-3).


Simplify this into whatever form or forms you need.
Find a good demonstration example for how this works, here: deriving equation of parabola for given focus and directrix - video example
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The finished general form for this example in your list is highlight%28y=%281%2F4%29x%5E2-2x%29.

Its graph appears like graph%28300%2C300%2C-1%2C9%2C-5%2C5%2Cx%5E2%2F4-2x%29.


The same channel with that video also has something to explain your last two example equations.