The equation of the tangent to the curve y = √(3x − 2) which is parallel to the line 3x − 4y + 1 = 0 is:
The equation of the tangent to the curve y = √(3x − 2) which is parallel to the line 3x − 4y + 1 = 0 is:
- A. 3x − 4y + 2 = 0
- B. 3x − 4y − 1 = 0
- C. 4x + 3y − 7 = 0
- D. 3x + 4y − 5 = 0
Answer: A) 3x − 4y + 2 = 0
Explanation: Slope of the given line = 3/4. dy/dx = 3/(2√(3x − 2)) = 3/4 ⇒ √(3x − 2) = 2 ⇒ x = 2, y = 2. Tangent: y − 2 = (3/4)(x − 2) ⇒ 3x − 4y + 2 = 0.
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