The graph of the conic x^2 - (y - 1)^2 = 1 ha | vedclass

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(D) The given equation of the conic is $x^2 - (y - 1)^2 = 1$. This is a hyperbola with center $(0, 1)$.Since the equation is of the form $\frac{X^2}{A

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$\sqrt{2}$

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(D) The given equation of the conic is $x^2 - (y - 1)^2 = 1$. This is a hyperbola with center $(0, 1)$.Since the equation is of the form $\frac{X^2}{A^2} - \frac{Y^2}{B^2} = 1$ where $A^2 = 1$ and $B^2 = 1$,it is a rectangular hyperbola.For a rectangular hyperbola,the eccentricity $e$ is given by $e = \sqrt{1 + \frac{B^2}{A^2}}$.Substituting $A^2 = 1$ and $B^2 = 1$,we get $e = \sqrt{1 + \frac{1}{1}} = \sqrt{2}$.Thus,the eccentricity of the conic is $\sqrt{2}$.

The graph of the conic $x^2 - (y - 1)^2 = 1$ has one tangent line with positive slope that passes through the origin. The point of tangency is $(a, b)$. Then the eccentricity of the conic is:

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