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(A) The equation of the ellipse is $x^2 + 4y^2 - 2x + 8y + 1 = 0$. Completing the square,we get $(x-1)^2 + 4(y+1)^2 = 4$,which simplifies to $\frac{(x

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lies inside the ellipse $S = 0$

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(A) The equation of the ellipse is $x^2 + 4y^2 - 2x + 8y + 1 = 0$. Completing the square,we get $(x-1)^2 + 4(y+1)^2 = 4$,which simplifies to $\frac{(x-1)^2}{4} + \frac{(y+1)^2}{1} = 1$. Let the tangent line passing through $(1, 1)$ be $y - 1 = m(x - 1)$,or $y = mx + (1 - m)$. For a line $y = mx + c$ to be a tangent to $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$,the condition is $(c - k + mh)^2 = a^2m^2 + b^2$. Here $h=1, k=-1, a^2=4, b^2=1$. So,$(1 - m - (-1) + m(1))^2 = 4m^2 + 1$. $(2)^2 = 4m^2 + 1 \Rightarrow 4 = 4m^2 + 1 \Rightarrow 4m^2 = 3 \Rightarrow m = \pm \frac{\sqrt{3}}{2}$. Given $m_1 > m_2$,we have $m_1 = \frac{\sqrt{3}}{2}$ and $m_2 = -\frac{\sqrt{3}}{2}$. The point $P$ is $(\frac{\sqrt{3}}{2}, -\frac{\sqrt{3}}{2})$. Substituting $P$ into the ellipse equation $S(x, y) = \frac{(x-1)^2}{4} + (y+1)^2 - 1$: $S(\frac{\sqrt{3}}{2}, -\frac{\sqrt{3}}{2}) = \frac{(\frac{\sqrt{3}}{2} - 1)^2}{4} + (-\frac{\sqrt{3}}{2} + 1)^2 - 1$. Since $(\frac{\sqrt{3}}{2} - 1)^2 = (1 - \frac{\sqrt{3}}{2})^2$,we have $S = \frac{(1 - \frac{\sqrt{3}}{2})^2}{4} + (1 - \frac{\sqrt{3}}{2})^2 - 1 = \frac{5}{4}(1 - \frac{\sqrt{3}}{2})^2 - 1$. Since $(1 - \frac{\sqrt{3}}{2})^2 \approx (1 - 0.866)^2 = (0.134)^2 \approx 0.018$,$S \approx \frac{5}{4}(0.018) - 1 Thus,the point lies inside the ellipse.

Tangents are drawn from point $(1, 1)$ to the ellipse $S \equiv x^2 + 4y^2 - 2x + 8y + 1 = 0$. If $m_1, m_2$ $(m_1 > m_2)$ are the slopes of these tangents,then with respect to the given ellipse,the point $P(m_1, m_2)$:

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