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

Finding intersection points:

The parabola $y^2 = 2x$ intersects the line $y = 2x$ where: $$(2x)^2 = 2x$$ $$4x^2 = 2x$$ $$4x^2 - 2x = 0$$ $$2x(2x - 1) = 0$$ $$x = 0 \text{ or } x = \frac{1}{2}$$

Intersection points: $(0, 0)$ and $\left(\frac{1}{2}, 1\right)$

Arc length formula:

For a curve, arc length can be computed using:

• If $x = g(y)$: $L = \int_{y_1}^{y_2} \sqrt{1 + \left(\frac{dx}{dy}\right)^2} dy$
• If $y = f(x)$: $L = \int_{x_1}^{x_2} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx$

Method 1: Parameterize with $y$

From $y^2 = 2x$, we get $x = \frac{y^2}{2}$

$$\frac{dx}{dy} = y$$

Arc length: $$L = \int_0^1 \sqrt{1 + y^2} , dy$$

This matches (A)

Method 2: Parameterize with $x$

From $y^2 = 2x$, we get $y = \sqrt{2x}$ (taking positive branch since $y = 2x$ is positive)

$$\frac{dy}{dx} = \frac{1}{2\sqrt{2x}} \cdot 2 = \frac{1}{\sqrt{2x}}$$

Arc length: $$L = \int_0^{1/2} \sqrt{1 + \left(\frac{1}{\sqrt{2x}}\right)^2} , dx$$

$$= \int_0^{1/2} \sqrt{1 + \frac{1}{2x}} , dx$$

$$= \int_0^{1/2} \sqrt{\frac{2x + 1}{2x}} , dx$$

$$= \int_0^{1/2} \frac{\sqrt{2x + 1}}{\sqrt{2x}} , dx$$

$$= \int_0^{1/2} \frac{\sqrt{1 + 2x}}{\sqrt{2x}} , dx$$

This matches (C) 

Verification that (A) and (C) are equivalent:

Let $y = \sqrt{2x}$, then $dy = \frac{1}{\sqrt{2x}} dx$

When $x = 0$: $y = 0$ When $x = \frac{1}{2}$: $y = 1$

$$\int_0^{1/2} \frac{\sqrt{1+2x}}{\sqrt{2x}} dx = \int_0^1 \sqrt{1 + y^2} , dy$$

Both expressions are equivalent.

Answer: (A) and (C) are correct