### Instruction: 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 $|ab|.$.### Response: I recognize these equations as representing ellipses and hyperbolas, respectively.
I recall that for an ellipse with major axis length $2a$ and minor axis length $2b,$ the distance from the center to each focus is $c,$ where $c^2 = a^2 - b^2.$
Similarly, for a hyperbola with transverse axis length $2a$ and conjugate axis length $2b,$ the distance from the center to each focus is $c,$ where $c^2 = a^2 + b^2.$
In both cases, the center is at the origin, since the equations are symmetric about both axes.
Since the foci of the ellipse are at $(0,\pm 4),$ I deduce that $c = 4$ for the ellipse, and hence $a^2 - b^2 = 16.$
Since the foci of the hyperbola are at $(\pm 6,0),$ I deduce that $c = 6$ for the hyperbola, and hence $a^2 + b^2 = 36.$
Adding these two equations, I get $2a^2 = 52,$ or $a^2 = 26.$
Subtracting these two equations, I get $2b^2 = 20,$ or $b^2 = 10.$