In Exercises 83–88, find equations for the lines that are tangent, and the lines that are normal, to the curve at the given point.
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x + √xy = 6, (4, 1)

Verified step by step guidance

1. First, understand that to find the tangent line to the curve at a given point, we need to determine the derivative of the curve with respect to x. The given curve is x + \sqrt{xy} = 6.
2. To find the derivative, use implicit differentiation. Differentiate both sides of the equation with respect to x. Remember that when differentiating \sqrt{xy}, you will need to use the chain rule and the product rule.
3. After differentiating, solve for \frac{dy}{dx}, which represents the slope of the tangent line at any point (x, y) on the curve.
4. Substitute the given point (4, 1) into the expression for \frac{dy}{dx} to find the slope of the tangent line at this specific point.
5. Once you have the slope, use the point-slope form of a line, y - y_1 = m(x - x_1), where m is the slope and (x_1, y_1) is the point (4, 1), to write the equation of the tangent line. For the normal line, use the negative reciprocal of the tangent slope and apply the same point-slope formula.