The area enclosed by y² = 8x and y = √2x that lies outside the triangle formed by y = √2x, x = 1, y

Q: Wait, how do I know if the triangle is actually inside the region enclosed by the parabola and the line?

Great question! Always check the boundaries. The line $y = \sqrt{2}x$ is a boundary for both the parabola region and the triangle. By sketching the graph, you will see the triangle sits snugly within the region defined by the parabola and the line. It is a classic JEE geometry check!

Q: Why can't I just use the formula for the area of a parabola segment directly?

You absolutely can! The area of a parabolic segment cut by a chord is $\frac{8}{3} a^2 m^3$ or similar variants, but for JEE, mastering the integral method is safer and more versatile. It builds the intuition you need for more complex problems where standard formulas do not apply.

Q: Is the triangle always a right-angled triangle in these types of problems?

Not always, but in this specific case, the boundaries $x=1$ (vertical) and $y=2\sqrt{2}$ (horizontal) naturally create a right angle. Always look for these geometric shortcuts—they save precious time during the exam!

Question

The area enclosed by $y^2 = 8x$ and $y = \sqrt{2}x$ that lies outside the triangle formed by $y = \sqrt{2}x, x = 1, y = 2\sqrt{2}$, is equal to :

Options:

(A) $\frac{16\sqrt{2}}{6}$

(B) $\frac{11\sqrt{2}}{6}$

(C) $\frac{13\sqrt{2}}{6}$

(D) $\frac{5\sqrt{2}}{6}$

Sigma Percentile · Accepted Answer

### The Geometry of Hidden Spaces

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving an integral; we are exploring the landscape of coordinate geometry.

Imagine you are standing on the Cartesian plane. To your right, the parabola $y^2 = 8x$ stretches out, a graceful curve opening towards infinity. Cutting through this curve is the line $y = \sqrt{2}x$.

Together, they enclose a beautiful, leaf-like region. Our task is to find the area of this region, excluding a specific triangle hidden within it. This is a journey of subtraction, of finding the whole and then carefully removing the part that does not belong.

### Phase 1: The Dance of Curves

Before we calculate, we must understand the boundaries. We find the intersection points by substituting $y = \sqrt{2}x$ into $y^2 = 8x$.

$$(\sqrt{2}x)^2 = 8x \Rightarrow 2x^2 = 8x$$

Solving the equation $2x(x - 4) = 0$, we find the intersection points at $x = 0$ and $x = 4$. These are the anchors of our region.

At $x = 0$, we are at the origin $(0, 0)$. At $x = 4$, we reach the point $(4, 4\sqrt{2})$. This interval defines the span of our integration.

### Phase 2: The Integral

Now, we calculate the total area enclosed by the parabola and the line. The area between two curves is the integral of the upper curve minus the lower curve.

Here, the parabola $y = \sqrt{8x}$ is the upper boundary, and the line $y = \sqrt{2}x$ is the lower boundary. We set up the integral as follows:

$$A_{total} = \int_{0}^{4} (\sqrt{8x} - \sqrt{2}x) dx$$

We integrate $2\sqrt{2}x^{1/2}$ to get $\frac{4\sqrt{2}}{3}x^{3/2}$, and we integrate $\sqrt{2}x$ to get $\frac{\sqrt{2}}{2}x^2$. Evaluating this from $0$ to $4$:

$$A_{total} = \left[ \frac{4\sqrt{2}}{3}x^{3/2} - \frac{\sqrt{2}}{2}x^2 ight]_{0}^{4} = \frac{32\sqrt{2}}{3} - 8\sqrt{2} = \frac{8\sqrt{2}}{3}$$

### Phase 3: The Triangle Trap

The problem introduces a triangle formed by the lines $y = \sqrt{2}x$, $x = 1$, and $y = 2\sqrt{2}$. Let us identify its vertices.

The intersection of $x = 1$ and $y = \sqrt{2}x$ gives $(1, \sqrt{2})$. The intersection of $x = 1$ and $y = 2\sqrt{2}$ gives $(1, 2\sqrt{2})$. Finally, the intersection of $y = 2\sqrt{2}$ and $y = \sqrt{2}x$ gives $x = 2$, resulting in the point $(2, 2\sqrt{2})$.

This is a right-angled triangle with a base of length $1$ (from $x=1$ to $x=2$) and a height of $\sqrt{2}$ (from $y=\sqrt{2}$ to $y=2\sqrt{2}$). Its area is:

$$A_{triangle} = \frac{1}{2} 	imes 1 	imes \sqrt{2} = \frac{\sqrt{2}}{2}$$

### The Final Revelation

We have the total area, $\frac{8\sqrt{2}}{3}$, and the area of the triangle, $\frac{\sqrt{2}}{2}$. To find the area outside the triangle, we perform the subtraction:

$$A_{final} = \frac{8\sqrt{2}}{3} - \frac{\sqrt{2}}{2}$$

Finding a common denominator of $6$, we obtain:

$$A_{final} = \frac{16\sqrt{2} - 3\sqrt{2}}{6} = \frac{13\sqrt{2}}{6}$$

The final result is **$\frac{13\sqrt{2}}{6}$**. You have navigated the curves, mastered the integral, and uncovered the hidden space.

Animated Solution for Mathematics - Definite Integration: The area enclosed by $y^{2} = 8 x$ and $y = 2 x$ that lies outside the triangle formed by $y = 2 x, x = 1, y = 22$ , is equal to :

Select Answer:

Visualized Solution

Visualizing the Curves $y^{2} = 8 x$ and $y = 2 x$

Finding Intersection Points

Solving for $x$

Setting up the Total Area Integral

Integrating the Area Function

Evaluating the Total Area

Identifying the Triangle Boundaries

Finding the Triangle Vertices

Calculating the Triangle Area

Finding the Final Enclosed Area

The Sigma Insight: Area Bounded by Curves

The Geometry of Hidden Spaces

Phase 1

Phase 2

Phase 3

The Final Revelation

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Animated Solution for Mathematics - Definite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The area enclosed by y2=8x and y=2​x that lies outside the triangle formed by y=2​x,x=1,y=22​, is equal to :