What is the equation of the quadratic graph with a focus of (1, -3) and a directrix of y= -1 ? show work hint: y=a(x-h)^2+k, a=1/4 (x-x1)^2+ 9y-y1)^2= (y-y2)^2

Step-by-step explanation:

Given : focus of (1, -3) and a directrix of y= -1

The distance between focus and directrix is 2p

The distance between (1,-3)  and y=-1  is 2

So 2p = 2  and p =1

Vertex (h,k) lies inbetween the focus and directrix

focus is (h, k-p)

given focus is (1,-3)

so h=1  and k-p = -3 ( we got p =1)

So k-1= -3  , k= -2

Hence, vertex is (1,-2), h=1  and k = -2

Here focus at the bottom and directrix at the top

so parabola opens down

hence a= -1/4

Plug it in the equation

{x = -1/2 ,   y = -1

step-by-step explanation:

solve the following system:

{2 x + y = -2

{10 x - 3 y = -2

in the first equation, look to solve for y:

{2 x + y = -2

{10 x - 3 y = -2

subtract 2 x from both sides:

{y = -2 x - 2

{10 x - 3 y = -2

substitute y = -2 x - 2 into the second equation:

{y = -2 x - 2

{10 x - 3 (-2 x - 2) = -2

10 x - 3 (-2 x - 2) = (6 x + 6) + 10 x = 16 x + 6:

{y = -2 x - 2

{16 x + 6 = -2

in the second equation, look to solve for x:

{y = -2 x - 2

{16 x + 6 = -2

subtract 6 from both sides:

{y = -2 x - 2

{16 x = -8

divide both sides by 16:

{y = -2 x - 2

{x = -1/2

substitute x = -1/2 into the first equation:

{y = -1

{x = -1/2

collect results in alphabetical order:

answer:   {x = -1/2 ,   y = -1