Final answer:
To find the coordinates of point P, we first find the equation of the normal to the curve at the point (1,11), which is y = x + 10. Then, we solve the equations y = 2x² - 5x + 14 and y = x + 10 simultaneously to find the coordinates of P. The coordinates of point P are (2, 12) and (1, 11).
Step-by-step explanation:
To find the coordinates of point P, we need to find the equation of the normal to the curve at the point (1,11) and then find the point of intersection between the normal and the curve. The slope of the tangent to the curve at any point is given by the derivative of the curve equation. Taking the derivative of y = 2x² - 5x + 14, we get dy/dx = 4x - 5. The slope of the normal is the negative reciprocal of the slope of the tangent, so the slope of the normal at (1,11) is -1/(4*1 - 5) = 1/1 = 1.
We know that the equation of a line with slope m passing through a point (x₁,y₁) is given by y - y₁ = m(x - x₁). Plugging in the values x₁ = 1, y₁ = 11, and m = 1, we get y - 11 = 1(x - 1), which simplifies to y = x + 10.
To find the coordinates of point P, we solve the equations y = 2x² - 5x + 14 and y = x + 10 simultaneously. Substituting y = x + 10 in the first equation, we get x + 10 = 2x² - 5x + 14. Rearranging this equation gives 2x² - 6x + 4 = 0.
To solve this quadratic equation, we can use the quadratic formula x = (-b ± √(b² - 4ac))/(2a), where a = 2, b = -6, and c = 4. Plugging in these values, we get x = (-(-6) ± √((-6)² - 4(2)(4)))/(2(2)), which simplifies to x = (6 ± √(36 - 32))/4, x = (6 ± √4)/4, x = (6 ± 2)/4. So x can be either 2 or 1. Plugging these values back into y = x + 10, we get the corresponding y-values as y = 2 + 10 = 12 and y = 1 + 10 = 11. Therefore, the coordinates of point P are (2, 12) and (1, 11).