Question

The point on the curve $y^2 = 2(x - 3)$ at which the normal is parallel to the line $y - 2x + 1 = 0$ is

• A. $(-\frac{1}{2}, -2)$
• B. $(\frac{3}{2}, 2)$
• C. $(5, 2)$
• D. $(5, -2)$

Hint:

To verify your result quickly in an exam, check if the calculated options satisfy the original curve equation. Substitution of option (C) and (D) coordinates both yield $4 = 4$, making them valid points on the curve. Then, check the sign of $y$: since the line has a positive slope ($+2$) and the normal is parallel to it, the tangent slope must be negative ($-\frac{1}{2}$). Since $\frac{dy}{dx} = \frac{1}{y}$, $y$ must be negative, eliminating option (C) instantly!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
Find the point on the parabola y² = 2(x - 3) where the normal is parallel to the line y - 2x + 1 = 0.

Step 2: Key Formula or Approach:
The given line has slope m = 2. Since the normal is parallel, m_normal = 2. Tangent slope m_tangent = -1/m_normal = -1/2. Find dy/dx implicitly from the curve, equate to m_tangent, and solve.

Step 3: Detailed Explanation:
Given line: y = 2x - 1 → slope = 2 → m_normal = 2 → m_tangent = -1/2. Differentiating y² = 2(x - 3): 2y(dy/dx) = 2 → dy/dx = 1/y. Setting 1/y = -1/2 → y = -2. Substitute into curve: (-2)² = 2(x - 3) → 4 = 2(x - 3) → x - 3 = 2 → x = 5. The point is (5, -2).

Step 4: Final Answer:
Coordinates are (5, -2), option (D).