To find the equation of the tangent and normal lines to the curve at a given point, we first need to find the derivative of the curve implicitly to get the slope of the tangent line at that point. The slope of the normal line is the negative reciprocal of the tangent slope. Using the point-slope form of a line, we can then write the equations for both the tangent and normal lines.
We start with the curve defined by the equation
\[
x^{2} y^{2} - 2x + 9y = 20.
\]
To find the slope of the tangent line at the point \((1, 2)\), we differentiate both sides implicitly with respect to \(x\):
\[
\frac{d}{dx}(x^{2} y^{2}) - \frac{d}{dx}(2x) + \frac{d}{dx}(9y) = 0.
\]
This gives us
\[
2xy^{2} + 2x^{2}y\frac{dy}{dx} - 2 + 9\frac{dy}{dx} = 0.
\]
Rearranging terms, we find
\[
\frac{dy}{dx} = \frac{2 - 2xy^{2}}{2x^{2}y + 9}.
\]
Next, we evaluate the derivative at the point \((1, 2)\):
\[
\frac{dy}{dx} \bigg|_{(1, 2)} = \frac{2 - 2(1)(2^{2})}{2(1^{2})(2) + 9} = \frac{2 - 8}{2 + 9} = \frac{-6}{11} = \frac{6}{13}.
\]
Thus, the slope of the tangent line at the point \((1, 2)\) is
\[
m_{\text{tangent}} = \frac{6}{13}.
\]
The slope of the normal line is the negative reciprocal of the tangent slope:
\[
m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}} = -\frac{13}{6}.
\]
Using the point-slope form of the line, we can write the equations for both the tangent and normal lines.
Tangent Line:
\[
y - 2 = \frac{6}{13}(x - 1).
\]
Rearranging gives:
\[
y = \frac{6}{13}x + \frac{20}{13}.
\]
Normal Line:
\[
y - 2 = -\frac{13}{6}(x - 1).
\]
Rearranging gives:
\[
y = -\frac{13}{6}x + \frac{25}{6}.
\]
The equations of the tangent and normal lines at the point \((1, 2)\) are:
Tangent Line: \(\boxed{y = \frac{6}{13}x + \frac{20}{13}}\)
Normal Line: \(\boxed{y = -\frac{13}{6}x + \frac{25}{6}}\)
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