Question

The arc length of the parabola $y^2 = 2x$ intercepted between the points of intersection of the parabola $y^2 = 2x$ and the straight line $y = 2x$ equals

• A. $\int_{0}^{1} \sqrt{1 + 4y^2} \, dy$
• B. $\int_{0}^{1} \sqrt{1 + 4y^2} \, dy$
• C. $\int_{0}^{1/2} \dfrac{\sqrt{1 + 4x}}{\sqrt{2x}} \, dx$
• D. $\int_{0}^{1/2} \dfrac{\sqrt{1 + 4x}}{\sqrt{2x}} \, dx$

Hint:

Always check which variable simplifies the derivative for the arc length integral—sometimes switching between $x$ and $y$ simplifies computation.

Answer 1

The Correct Option is A, C

Step 1: Points of intersection. 
Given $y^2 = 2x$ and $y = 2x$. Substitute $x = y/2$ into $y^2 = 2x$: \[ y^2 = 2 \times \frac{y}{2} \Rightarrow y^2 = y \Rightarrow y = 0 \text{ or } y = 1. \] Thus, $x = 0$ and $x = \frac{1}{2}$.

Step 2: Arc length formula for a curve $y^2 = 2x$. 
For $y^2 = 2x$, we can express $x = \frac{y^2}{2}$. Then, \[ \frac{dx}{dy} = y. \] Arc length between $y = 0$ and $y = 1$ is \[ L = \int_{0}^{1} \sqrt{1 + \left(\frac{dx}{dy}\right)^2} \, dy = \int_{0}^{1} \sqrt{1 + y^2} \, dy. \] However, using $x$ as parameter: $y = \sqrt{2x}$, $\dfrac{dy}{dx} = \dfrac{1}{\sqrt{2x}}$.

Step 3: Arc length in terms of $x$. 
\[ L = \int_{0}^{1/2} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx = \int_{0}^{1/2} \sqrt{1 + \frac{1}{2x}} \, dx = \int_{0}^{1/2} \frac{\sqrt{1 + 4x}}{\sqrt{2x}} \, dx. \]

Step 4: Conclusion. 
\[ \boxed{L = \int_{0}^{1/2} \frac{\sqrt{1 + 4x}}{\sqrt{2x}} \, dx.} \]