Question

Write a standard equation of the parabola with focus (2,1) and vertex (-5,1)

Answer 1

so, the vertex is at (-5, 1) and the focus is at (2, 1)
notice the "y" didn't change, the "x" did
that means the parabola is opening towards the focus, that is, to the right-handside as show above

Answer 2

so what would the equation be for that in standard form

Answer 3

$$
(y-\color{blue}{k})^2 = 4p(x-\color{red}{h})\\
\text{"p" is the distance from the vertex to the focus}\\
\text{from -5 to +2, you have 7 units, thus p = 7}\\
\text{the vertex is at} (\color{red}{h}, \color{blue}{k})
$$

Answer 4

so, you have already (h, k) = (-5, 1)
and p = 7

Answer 5

Thank you so much and do uhappen to know how to do this And do you happen to know how you would write this equation into a calculater y^2-4x+2y+5=0

Answer 6

sadly, most calculators use function based on "y" for their plotting, so for y^2-4x+2y+5=0, you'd need to solve for "y", which can get cumbersome I just found out hehe

Answer 7

but what you can do is, make that into a (y-k)^2 = 4p(x-h) form, gimme a few secs

Answer 8

\(y^2+2y = 4x-5\)
# completing the square for the perfect trinomial on the left-hand side
\((y^2+2y+1^2) = 4x-5\)
# getting the perfect square trinomial, and grouping on the right-side
\((y+1)^2 = 4(x-\frac{5}{4})\)

Answer 9

so, your vertex is at (5/4, -1)

to plot the parabola opening, it has a y^2 component, meaning is opens horizontally
4P = 4, meaning P = 1, is a positive value, meaning it opens to the right-hand side

so just pick any 2 points greater than 5/4 for "x" and solve for "y"

Answer 10

so, when say x = 3
\((y+1)^2 = 7, y = \sqrt{7} -1 = 1.645\)

Answer 11

though once you have the equation above, you can just also give a calculator
\((y+1)^2 = 4\pmatrix{x-\frac{5}{4}} \implies
y = \sqrt{4\pmatrix{x-\frac{5}{4}}}-1 \)

Answer 12

I just notice one tiny mistake I made, when doing the completing of the square, I was supposed to do -1 to the left side, so
the correct equation will be
$$\large{
(y+1)^2 = 4\pmatrix{x-\frac{6}{4}} \implies
y = \sqrt{4\pmatrix{x-\frac{3}{2}}}-1
}
$$