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

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(D) The equation of the hyperbola is $x^2 - (y - 1)^2 = 1$. Let the tangent line passing through the origin $(0, 0)$ be $y = mx$. Since the line is ta

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$\frac{\pi}{4}$

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(D) The equation of the hyperbola is $x^2 - (y - 1)^2 = 1$. Let the tangent line passing through the origin $(0, 0)$ be $y = mx$. Since the line is tangent to the hyperbola,the perpendicular distance from the center $(0, 1)$ to the line $mx - y = 0$ must be equal to the semi-transverse axis $a = 1$. Using the formula for distance from a point $(x_1, y_1)$ to a line $Ax + By + C = 0$,$d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$. $1 = \frac{|m(0) - 1(1)|}{\sqrt{m^2 + (-1)^2}}$ $\Rightarrow 1 = \frac{1}{\sqrt{m^2 + 1}}$ $\Rightarrow m^2 + 1 = 1$ $\Rightarrow m = 0$. Wait,the slope must be positive. Let's re-evaluate. The tangent line is $y = mx$. The point of tangency $(a, b)$ satisfies $b = ma$ and $a^2 - (b - 1)^2 = 1$. Differentiating $x^2 - (y - 1)^2 = 1$ with respect to $x$: $2x - 2(y - 1)\frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = \frac{x}{y - 1}$. At $(a, b)$,the slope is $m = \frac{a}{b - 1}$. Since $m = \frac{b}{a}$,we have $\frac{b}{a} = \frac{a}{b - 1} \Rightarrow a^2 = b^2 - b$. Substituting $a^2 = b^2 - b$ into the hyperbola equation $a^2 - (b - 1)^2 = 1$: $(b^2 - b) - (b^2 - 2b + 1) = 1$ $\Rightarrow b^2 - b - b^2 + 2b - 1 = 1$ $\Rightarrow b - 1 = 1$ $\Rightarrow b = 2$. Then $a^2 = 2^2 - 2 = 2 \Rightarrow a = \sqrt{2}$ (since the point is in the first quadrant). Thus,$\sin^{-1}\left(\frac{a}{b}\right) = \sin^{-1}\left(\frac{\sqrt{2}}{2}\right) = \sin^{-1}\left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4}$.

The graph of the conic $x^2-(y-1)^2=1$ has one tangent line with positive slope that passes through the origin. If the point of tangency is $(a, b)$,then find the value of $\sin^{-1}\left(\frac{a}{b}\right)$.

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