This is what I have so far, but I’m not sure what I need to fix. please help asap!!

I think your answer is correct already. You just have to put your latus rectum, directrix, and focus into your drawing. :)

for asymptotes:

Let's change the centers into (-3, 2). Therefore, (y-2)² - (x+3)²/16 = 1

To get the asymptotes, let's equate the equation to zero (0) instead of one (1).

(y-2)² - (x+3)²/16 = 0

Notice that this equation is the difference of two squares.

Let x² = (y-2)² and y² = (x+3)²/16

Therefore, we have x² - y² =0

The factors of two squares are:

(x+y)(x-y)=0

Then, equating each factor to zero:

x+y=0

x-y=0

Substituting the real equations:

» (y-2) + (x+3)/4 = 0

y-2 = - (x+3)/4

y= -(x+3)/4 + 2

» (y-2) - (x+3)/4 = 0

y-2 = (x+3)/4

y= (x+3)/4 +2

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3