Final answer:
The equation of the tangent line at the point (9, 27) on the curve y = x² is y = 18x - 135, and the equation of the normal line is y = (-1/18)x + 27.5.
Step-by-step explanation:
To find the equations of the tangent line and normal line to the curve y = x² at the point (9, 27), first we need to determine the slope of the tangent line at that point. The derivative of y with respect to x gives us the slope of the tangent to the curve at any point x. So we calculate the derivative dy/dx:
dy/dx = 2x.
At the point (9, 27), the slope is 2 × 9 = 18. Therefore, the equation of the tangent line, using the point-slope form, is:
y - 27 = 18(x - 9).
Simplifying, we find the equation of the tangent line is y = 18x - 135.
For the normal line, which is perpendicular to the tangent, we take the negative reciprocal of the tangent's slope. This gives us a slope of -1/18. Using the point-slope form again:
y - 27 = (-1/18)(x - 9).
Simplifying, we find the equation of the normal line is y = (-1/18)x + 27 + 1/2 or y = (-1/18)x + 27.5.