The equation frac{x^2}{1 - r} - frac{y^2}{1 + | vedclass

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(D) Given the equation: $\frac{x^2}{1 - r} - \frac{y^2}{1 + r} = 1$ where $r > 1$.Since $r > 1$,let $r - 1 = p$,where $p > 0$. Then $1 - r = -p$.Subst

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An imaginary ellipse

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(D) Given the equation: $\frac{x^2}{1 - r} - \frac{y^2}{1 + r} = 1$ where $r > 1$.Since $r > 1$,let $r - 1 = p$,where $p > 0$. Then $1 - r = -p$.Substituting this into the equation,we get: $\frac{x^2}{-p} - \frac{y^2}{1 + r} = 1$.Multiplying by $-1$,we get: $\frac{x^2}{p} + \frac{y^2}{1 + r} = -1$.Since the sum of two squares divided by positive constants equals a negative value $(-1)$,there are no real values of $(x, y)$ that satisfy this equation.Therefore,the equation represents an imaginary ellipse.

The equation $\frac{x^2}{1 - r} - \frac{y^2}{1 + r} = 1$ for $r > 1$ represents:

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