To find the equation of a line perpendicular to the given equation and passing through the point \((-4, -2)\), we first need to rewrite the given equation \( y + 2 = \frac{1}{2}xy + 2 \) in a more manageable form.
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Rearrange the given equation:
\[ y + 2 = \frac{1}{2}xy + 2 \]
Subtract 2 from both sides:
\[ y = \frac{1}{2}xy \]
Rearranging this gives:
\[ y - \frac{1}{2}xy = 0 \]
Factoring out \(y\):
\[ y(1 - \frac{1}{2}x) = 0 \]
This results in two equations: \(y = 0\) and \(1 - \frac{1}{2}x = 0\), but since we are looking for a slope, focusing on \(1 - \frac{1}{2}x = 0\) gives:
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Solve for \(y\):
Rearranging gives:
\[ y = \frac{2}{x} \]
The slope of this curve when expressed in general form is not constant, but to find the perpendicular line, we need the slope of the linear approximation or evaluation at a point.
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Find the slope of a tangent line:
To find the slope at \(\left(-4, -2\right)\), we can take the derivative of \(y\):
\[ \frac{dy}{dx} = -\frac{2}{x^2} \]
Substitute \(x = -4\):
\[ \frac{dy}{dx} \bigg|_{x=-4} = -\frac{2}{(-4)^2} = -\frac{2}{16} = -\frac{1}{8} \]
If the slope of the tangent line is \( -\frac{1}{8} \), then the slope of a line perpendicular to this line is the negative reciprocal:
\[ m_{\text{perpendicular}} = -\frac{1}{\left(-\frac{1}{8}\right)} = 8 \]
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Now that we have the slope of the perpendicular line, use point-slope form to find the equation of the line that passes through the point \((-4, -2)\):
\[ y - y_1 = m(x - x_1) \]
Here, \(m = 8\), \(x_1 = -4\), and \(y_1 = -2\):
\[ y - (-2) = 8(x - (-4)) \]
Simplifying this gives:
\[ y + 2 = 8(x + 4) \]
\[ y + 2 = 8x + 32 \]
\[ y = 8x + 30 \]
Thus, the equation of the line perpendicular to the given equation and passing through the point \((-4, -2)\) is:
\[ \boxed{y = 8x + 30} \]