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
+118609
All the info you need is contained in these 2 diagrams.
Incorrect graph has been deleted : sorry about that.
+118609 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
