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:
We need to find a specific coordinate point on the parabola $y^2 = 2(x - 3)$ where the normal line is perfectly parallel to a given linear line $y - 2x + 1 = 0$.

Step 2: Key Formula or Approach:
1. Find the slope of the given line by converting it to slope-intercept form ($y = mx + c$). 2. Since the normal is parallel to this line, their slopes are equal ($m_{\text{normal}} = m_{\text{line}}$). 3. Use the relationship between the tangent and normal slopes: $m_{\text{tangent}} \times m_{\text{normal}} = -1 \implies m_{\text{tangent}} = -\frac{1}{m_{\text{normal}}}$. 4. Differentiate the curve equation implicitly to get an expression for $\frac{dy}{dx}$ (which is $m_{\text{tangent}}$) and solve for the coordinates.

Step 3: Detailed Explanation:
The equation of the given line is: $$y - 2x + 1 = 0 \implies y = 2x - 1$$ Thus, the slope of this line is $m = 2$. Since the normal is parallel to this line, the slope of the normal is: $$m_{\text{normal}} = 2$$ The slope of the tangent at any point is perpendicular to the normal line: $$m_{\text{tangent}} = -\frac{1}{m_{\text{normal}}} = -\frac{1}{2}$$ Now, let's find the derivative of the curve $y^2 = 2(x - 3)$ by differentiating implicitly with respect to $x$: $$2y \frac{dy}{dx} = 2(1 - 0)$$ $$2y \frac{dy}{dx} = 2 \implies \frac{dy}{dx} = \frac{1}{y}$$ Since $\frac{dy}{dx}$ represents the slope of the tangent line, we set it equal to $-\frac{1}{2}$: $$\frac{1}{y} = -\frac{1}{2} \implies y = -2$$ Substitute $y = -2$ back into the curve equation to compute the corresponding $x$-coordinate: $$(-2)^2 = 2(x - 3)$$ $$4 = 2(x - 3)$$ $$2 = x - 3 \implies x = 5$$ The coordinates of the required point are $(5, -2)$.

Step 4: Final Answer:
The coordinates of the point on the curve are $(5, -2)$, matching option (D).