Final answer:
The equations for the tangent and normal lines at point P(4, -3) on the given curve are Tangent: y = 3x - 15 and Normal: y = -3x + 15, which corresponds to option (c).
Step-by-step explanation:
To find the equations of the tangent and normal lines at a point on a given curve, we first need to derive the slope of the tangent line at that point by finding the first derivative of the equation of the curve. The equation of the curve provided is -x²y² + 2x³ = 2x - y - 27. After differentiating and evaluating the derivative at point P(4, -3), we get the slope of the tangent line, mt. The slope of the normal line, mn, is the negative reciprocal of mt since normal lines are perpendicular to tangent lines. Using the slope-point form (y - y1 = m(x - x1)) for both the tangent line and the normal line, we can write their equations. After simplifying the equations considering the given point P(4, -3), we identify the correct answer among the given options.
The correct equations for the tangent and normal lines at point P are: Tangent: y = 3x - 15, Normal: y = -3x + 15.
Therefore, the correct answer is (c) Tangent: y = 3x - 15, Normal: y = -3x + 15.