+96 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 $|ab|.$
+129899 \(\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
+1713 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:
\(has\ its\ foci\ at\)
\(tommarvoloriddle \)
