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Question:

Let C be a curve given by y(x) = 1 + √(4x - 34), x > 34. If P is a point on C such that the tangent at P has slope 2/3, then a point through which the normal at P passes is: (3, -34), (4, -29), (2, 3), (1, 7)

(3, -34)

(4, -29)

(1, 7)

(2, 3)

Solution:

dy/dx = 1/(2√(4x - 34)) × 4 = 2/3 ⇒ 4x - 34 = 9 ⇒ x = 3
So, y = 4
Equation of normal at P(3, 4) is y - 4 = -3/2(x - 3)
i.e. 2y - 8 = -3x + 9 ⇒ 3x + 2y - 17 = 0