How to find the standard form of the equation of the hyperbola?

If the center is (h, k) and the vertices are on the same vertical line, the equation is

(y - k)^2 / a^2 - (x - h)^2 / b^2 = 1

where a = the distance from the center to either vertex

and √(a^2 + b^2) = the distance from the center to a focus

If the vertices are on the same horizontal line, the (x - h)^2 comes first over a^2 then minus (y - k)^2 / b^2 = 1

In your first, (0, 2) and (0, -2) are on the same vertical line (the y axis).

a = 2 making a^2 = 4, and √(4 + b^2) = 4 making 4 + b^2 = 16 and b^2 = 12

so y^2 / 4 - x^2 / 12 = 1

Then (2,0) and (6,0) are on the same horizontal line (the x axis).

You know the center is equidistant from the vertices so it's (4,0).

That makes a = 2 again and a^2 = 4.

and √(4 + b^2) = 4 again so b^2 = 12

so, (x - 4)^2 / 4 + y^2 / 12 = 1

It has a horizontal transverse axis if the vertices are on the same horizontal line, and a vertical transverse if the vertices are on the same vertical line. Also, if the y^2 term is positive and x^2 is negative, it's a vertical TA but if x^2 is positive and y^2 is negative, horizontal TA

you ought to end the sq. for the two x and y 4x^2 - 8x ....+y^2 + 3y ..... = 19 4(x^2 - 2x ......) + y^2 + 3y .......=19 discover the extensive type needed = (b/2)^2 FOR x pick +a million for y pick +9/4... stability equation collectively as putting in blamks 4(x^2-2x+ a million) + y^2 + 3y + 9/4 = 19 ( upload to the two aspects)+4+9/4 4(x-a million)^2 + (y+3/2)^2 = 19+4+(9/4) it particularly interior reason no longer A CIRCLE!!!!! that's an ellipse! Divide by utilising 4 (x-a million)^2 + (y+3/2)^2 / 4 = one hundred and one/sixteen OR conventional variety for ellipse: (x-a million)^2 + (y+3/2)^2 = one hundred and one ____________________ . . . . . . . . . 2^2..........4^2

(x^2/a^2) - (y^2/b^2) = 1 is horizontal

(y^2/a^2) - (x^2/b^2} = 1 is vertical

Foci: (+ or - c, 0) for horizontal

(0, + or - c) for vertical

c^2= a^2 + b^2

Hope this helps