Discussion
The region in which the partial differential equation
\[ (1 - x^{2})u_{xx} - u_{yy} = 0 \]
is an ellipse if :
Answer: C
The equation is elliptic when |x| > 1.
For \(Au_{xx} + Bu_{xy} + Cu_{yy} = 0\), the type depends on \(B^{2} - 4AC\): negative means elliptic. Here \(A = 1 - x^{2}\), \(B = 0\), \(C = -1\):
\[ \begin{aligned} B^{2} - 4AC &= 0 - 4(1 - x^{2})(-1) \\ &= 4(1 - x^{2}) \end{aligned} \]
This is negative when \(x^{2} \gt 1\), that is \(|x| \gt 1\). For \(|x| \lt 1\) it is positive and the equation is hyperbolic, and at \(|x| = 1\) it is parabolic. 'x > 1' and 'x < 1' each describe only part of, or the wrong part of, the line.
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