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

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(C) The given equation is $x^2 - (y - 1)^2 = 1$,which is a hyperbola centered at $(0, 1)$.Differentiating with respect to $x$: $2x - 2(y - 1)\frac{dy}

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$2$

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(C) The given equation is $x^2 - (y - 1)^2 = 1$,which is a hyperbola centered at $(0, 1)$.Differentiating with respect to $x$: $2x - 2(y - 1)\frac{dy}{dx} = 0$,so $\frac{dy}{dx} = \frac{x}{y - 1}$.Since the tangent passes through the origin $(0, 0)$ and the point of tangency $(a, b)$,the slope of the tangent is $\frac{b - 0}{a - 0} = \frac{b}{a}$.Equating the slopes: $\frac{b}{a} = \frac{a}{b - 1} \Rightarrow a^2 = b(b - 1) = b^2 - b$. $(1)$Since $(a, b)$ lies on the hyperbola: $a^2 - (b - 1)^2 = 1$ $\Rightarrow a^2 - (b^2 - 2b + 1) = 1$ $\Rightarrow a^2 - b^2 + 2b = 2$. $(2)$Substituting $(1)$ into $(2)$: $(b^2 - b) - b^2 + 2b = 2 \Rightarrow b = 2$.Then $a^2 = 2^2 - 2 = 2$,so $a = \sqrt{2}$ (since the slope is positive).The standard form of the hyperbola is $\frac{x^2}{1^2} - \frac{(y - 1)^2}{1^2} = 1$,where $a^2 = 1$ and $b^2 = 1$.The length of the latus rectum for a hyperbola $\frac{X^2}{A^2} - \frac{Y^2}{B^2} = 1$ is $\frac{2B^2}{A}$.Here $A = 1$ and $B = 1$,so the length of the latus rectum is $\frac{2(1)^2}{1} = 2$.

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

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