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 $(\pm 6,0).$ Compute the value of $|ab|.$

Pushy Jul 10, 2019

#1**+2 **

\(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\]has its foci at $(0,\pm 4) \)

\(\frac{x^2}{a^2}-\frac{y^2}{b^2} = 1\]has its foci at $(\pm 6,0).\)

Find l ab l

The first equation is an ellipse with its center at (0, 0) and its major axis along y

The equation for the focus is (0 , 0 ± c) where c = ±4

And we have that a^2 - b^2 = c^2 ⇒ a^2 - b^2 = 16 (1)

The second equation is a hyperbola with its center at (0, 0) and its major axis along x

The equation for the focus is (0 ± c, 0) where c = ±6

And we have that a^2 + b^2 = c^2 ⇒ a^2 + b^2 = 36 (2)

Add (1) and (2) and we have that

2a^2 = 52

a^2 = 26 ⇒ a = ±√26

And using (2)

26 + b^2 = 36

b^2 = 10 ⇒ b = ±√10

So

l ab l = l √26 * √10 l = l √ 2 * √13 * √2 * √5 l = 2√65

CPhill Jul 10, 2019

#2**+6 **

Sorry, CPhill, just wondering if you knew how to space in latex.

If you do not know, you would put a \ and a space at the end of each word in the latex section, hence,

has\ its\ foci\ at

which would equal to:

\(has\ its\ foci\ at\)

__\(tommarvoloriddle \)__

tommarvoloriddle
Jul 11, 2019