The graph of \[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\]

has its foci at $(0,\pm 4),$

while the graph of \[\frac{x^2}{a^2}-\frac{y^2}{b^2} = 1\]

has its foci at $(\pm 6,0).$

Find a and b

Guest Jan 13, 2021

#1**+2 **

All the info you need is contained in these 2 diagrams.

Incorrect graph has been deleted : sorry about that.

Melody Jan 13, 2021

#4**+1 **

From the circle I am getting

16 = a^{2} - b^{2}

and from the hyperbola I am getting

36 = a^{2} + b^{2}

solving these 2 equations gets

20 = 2b^{2}

b = \(\sqrt{10}\)

a = \(\sqrt{26}\)

however when I try to graph it, it looks nothing like your graph

https://www.desmos.com/calculator/qeszzqvt5h

Guest Jan 15, 2021

#8**+1 **

**I am glad you have been persistent.**

**Keep it up. **

Your answer is closer than mine.

I don't have my earlier working any more. It beats me where I got those numbers from. Sorry.

Let me start over.

The equation of an ellipse is \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)

focal length c = 4

If the major axis is horizonal then a > b

If the major axis is vertical then a < b

Since this one has a major axis of y=0 it is vertical so a is smaller than b

The focal length is c where

\(c^2=|a^2-b^2|\)

since a < b,

\(c^2=b^2-a^2\\ 16=b^2-a^2\\ \)

For the hyperbola

\(c^2=a^2+b^2\\ 36=a^2+b^2\\\)

solve them simultaneously and you get \(b=\sqrt{26}\qquad a=\sqrt{10}\)

Here is the graph

https://www.desmos.com/calculator/4wam1hbltm

Melody Jan 16, 2021