#1**+2 **

Don't see how this is possible

The center would have to be (0,0) since the foci are (0, ±10) = (0, ±c)

The hyperbola opens up/down and has the form

y^2/a^2 - x^2/b^2 = 1

The equation for the asymptotes is y = ±(a/b) x

So......c = the distance from the center to the foci = sqrt [ a^2 + b^2 ] = sqrt [ 3^2 + 4^2] = sqrt [ 25] = 5

So....the foci should be at (0, ±5)

And the vertices are (0, ±3)

And the equation is

y^2/ 3^2 - x^2/4^2 =1 → y^2/9 - x^2/16 = 1

Here's a graph : https://www.desmos.com/calculator/ciaausrfvm

CPhill Dec 10, 2020

#2**-1 **

EIGHT questions posted already today, Cdortch. If you REALLY don't understand how to attempt ANY of them or ANY parts of them, you are in a class too dvanced for your current skill level...... you need to go back and review/study your assignments to get a grasp of the concepts before posting all of your homework online. We are not here to do your homework for you. We are here to assist you with occasional gaps in understanding that you may have...but you should attemt to answer the questions on your own.....

Guest Dec 10, 2020