If the area (in sq. units) bounded by the parabola y² = 4λx and the line y = λx, λ 0, is 1/9, then

Q: Wait, why do we use $y = \sqrt{4\lambda x}$? Isn't it $\pm$?

Great observation! The parabola $y^2 = 4\lambda x$ is indeed symmetric about the $x$-axis. However, the line $y = \lambda x$ (with $\lambda > 0$) lies entirely in the first and third quadrants. The bounded region is trapped in the first quadrant, where $y$ is positive. Thus, we only consider the positive branch $y = 2\sqrt{\lambda}\sqrt{x}$.

Q: Can we integrate with respect to $y$ instead of $x$?

Absolutely! You could express $x$ in terms of $y$ for both curves ($x = y^2/4\lambda$ and $x = y/\lambda$) and integrate with respect to $y$. It is a fantastic way to verify your answer and build flexibility in your problem-solving toolkit.

Q: Why is the area $\frac{8}{3\lambda}$? Is there a shortcut?

Yes, there is a standard result for the area between a parabola $y^2 = 4ax$ and a line $y = mx$. The area is $\frac{8a^3}{3m^3}$. Here, $4a = 4\lambda \implies a = \lambda$, and $m = \lambda$. So, Area = $\frac{8\lambda^3}{3\lambda^3} = \frac{8}{3\lambda}$. It is a powerful tool for competitive exams, but always understand the derivation first!

Question

If the area (in sq. units) bounded by the parabola $y^2 = 4\lambda x$ and the line $y = \lambda x, \lambda > 0$, is $1/9$, then $\lambda$ is equal to :

Options:

(A) 24

(B) 48

(C) $4\sqrt{3}$

(D) $2\sqrt{6}$

Sigma Percentile · Accepted Answer

### Analyzing the Geometry

Imagine you are standing on a coordinate plane, looking at two distinct paths. One is a parabola, $y^2 = 4\lambda x$, a curve that gracefully opens to the right, expanding as it travels.

The other is a straight line, $y = \lambda x$, a rigid, unwavering path passing directly through the origin. These two paths trap a sliver of space between them, and our mission is to find the value of the constant $\lambda$ that makes this trapped area exactly $1/9$.

### Finding the Intersection

To measure this area, we must first know where the dance begins and ends. We need the limits of integration.

By substituting the line equation $y = \lambda x$ into the parabola equation $y^2 = 4\lambda x$, we get:

$$(\lambda x)^2 = 4\lambda x$$

This simplifies to $\lambda^2 x^2 - 4\lambda x = 0$. Factoring this, we find $\lambda x(\lambda x - 4) = 0$.

Since $\lambda > 0$, our intersection points are at $x = 0$ and $x = 4/\lambda$. These are our boundaries.

### The Integral

Now, we set up the integral. The area is the integral from $0$ to $4/\lambda$ of the upper curve minus the lower curve.

The parabola is the upper curve, so $y = 2\sqrt{\lambda}\sqrt{x}$. The line is the lower curve, $y = \lambda x$. Our integral is:

$$\int_{0}^{4/\lambda} (2\sqrt{\lambda}x^{1/2} - \lambda x) \, dx$$

This is a standard polynomial integral. Integrating the first term, $2\sqrt{\lambda}x^{1/2}$, gives us $\frac{4}{3}\sqrt{\lambda}x^{3/2}$. Integrating the second term, $\lambda x$, gives us $\frac{\lambda x^2}{2}$.

### The Final Calculation

Evaluating this from $0$ to $4/\lambda$, we substitute the upper limit. The first term becomes:

$$\frac{4}{3}\sqrt{\lambda}\left(\frac{4}{\lambda}\right)^{3/2} = \frac{32}{3\lambda}$$

The second term becomes:

$$\frac{\lambda}{2}\left(\frac{4}{\lambda}\right)^2 = \frac{8}{\lambda}$$

Subtracting these, we get:

$$\frac{32}{3\lambda} - \frac{24}{3\lambda} = \frac{8}{3\lambda}$$

Finally, we equate this to the given area:

$$\frac{8}{3\lambda} = \frac{1}{9}$$

Solving for $\lambda$, we find $3\lambda = 72$, which means **$\lambda = 24$**. You have successfully navigated the curves and conquered the integral.

Animated Solution for Mathematics - Definite Integration: If the area (in sq. units) bounded by the parabola $y^{2} = 4 λ x$ and the line $y = λ x, λ > 0$ , is $1/9$ , then $λ$ is equal to :

Select Answer:

Visualized Solution

Visualizing the Curves

Finding Intersection Points

Solving for $x$

Limits of Integration

Setting up the Area Integral

The Integral Expression

Integrating the First Term

Integrating the Second Term

Applying the Limits

Simplifying the First Term

Simplifying the Second Term

Calculating Net Area

Final Value of $λ$

The Sigma Insight: Area Bounded by Curves

Analyzing the Geometry

Finding the Intersection

The Integral

The Final Calculation

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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; } If the area (in sq. units) bounded by the parabola y2=4λx and the line y=λx,λ>0, is 1/9, then λ is equal to :