The area of the region, inside the ellipse x^2 + 4y^2 = 4 and outside the region bounded by the curves y = |x| - 1 and y = 1

The area of the region, inside the ellipse x² + 4y² = 4 and outside the region bounded by the curves y = |x| - 1 and y = 1 - |x|, is :

Q. The area of the region, inside the ellipse $x^2 + 4y^2 = 4$ and outside the region bounded by the curves $y = |x| - 1$ and $y = 1 - |x|$, is :

(A) $2\pi - 1$

(B) $3(\pi - 1)$

(C) $2(\pi - 1)$

(D) $2\pi - \frac{1}{2}$

Correct Answer: (C) $2(\pi - 1)$

The required area is obtained by subtracting the area of the square formed by modulus curves from the total area of the ellipse.

Step-by-Step Detailed Solution

Step 1: Standardize the Ellipse Equation

The given equation of the ellipse is $x^2 + 4y^2 = 4$. Dividing the entire equation by 4, we get:

\frac{x^2}{4} + \frac{y^2}{1} = 1\) Comparing with \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), we have: \(a^2 = 4 \Rightarrow a = 2\) \(b^2 = 1 \Rightarrow b = 1

Step 2: Calculate Total Area of the Ellipse

The area of an ellipse is given by the formula $A = \pi ab$. Substituting our values:

\text{Area of Ellipse} = \pi(2)(1) = 2\pi

Step 3: Analyze the Modulus Curves

The inner region is bounded by two curves:
1. $y = |x| - 1$
2. $y = 1 - |x|$

Let's break down $y = |x| - 1$:
- If $x \ge 0$, $y = x - 1$
- If $x < 0$, $y = -x - 1$

Let's break down $y = 1 - |x|$:
- If $x \ge 0$, $y = 1 - x$
- If $x < 0$, $y = 1 + x$

Step 4: Identify the Shape Formed by the Curves

The four lines bounding the inner region are:
$L_1: x - y = 1$
$L_2: x + y = -1$
$L_3: x + y = 1$
$L_4: -x + y = 1$

The vertices of this region are $(1, 0), (0, 1), (-1, 0), (0, -1)$. This forms a square (or a rhombus) centered at the origin.

Step 5: Calculate Area of the Inner Square

The area of a square with vertices on the axes at distance 'd' from origin is $2d^2$, or simply calculate the area of 4 identical triangles in each quadrant:

\text{Area of Inner Region} = 4 \times \int_{0}^{1} (1 - x) dx\) \(\text{Area} = 4 \left[ x - \frac{x^2}{2} \right]_{0}^{1}\) \(\text{Area} = 4 \left( 1 - \frac{1}{2} \right) = 4 \left( \frac{1}{2} \right) = 2

Step 6: Final Required Area

The question asks for the area inside the ellipse but outside the modulus region:

\text{Required Area} = \text{Total Area of Ellipse} - \text{Area of Inner Square}\) \(\text{Required Area} = 2\pi - 2\) \(\text{Required Area} = 2(\pi - 1)

Related Theory: Area Under Curves and Conics

The concept of "Area Under Curves" (Application of Integrals) is a cornerstone of the JEE Mathematics syllabus. It combines Coordinate Geometry with Integral Calculus to measure the size of planar regions bounded by functions.

1. The Geometry of the Ellipse

An ellipse is the locus of a point such that the sum of its distances from two fixed points (foci) is constant. In its standard form:

\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1

The area formula $A = \pi ab$ can be derived using integration by expressing $y$ in terms of $x$: $y = b \sqrt{1 - \frac{x^2}{a^2}}$ and integrating from $-a$ to $a$ (doubling the result for the full ellipse).

2. Understanding Modulus Curves

Modulus functions $|x|$ introduce "corners" or "v-shapes" in graphs. When we see $y = 1 - |x|$, it is essential to recognize the symmetry. Because the function remains unchanged when $x$ is replaced by $-x$, the graph is symmetric about the y-axis. In this problem, the intersection of two such modulus graphs creates a closed polygon. Recognizing vertices early ($x=0$ and $y=0$ intercepts) often bypasses the need for complex integration.

3. "Inside and Outside" Logic

In JEE problems, the wording "Inside A and Outside B" mathematically translates to: $$\text{Area} = \text{Area}(A) - \text{Area}(A \cap B)$$ If B is entirely contained within A, it simply becomes $\text{Area}(A) - \text{Area}(B)$. Before integrating, always check if the boundaries of region B lie within the boundaries of region A. In our case, the square vertices $(\pm 1, 0)$ and $(0, \pm 1)$ are well within the ellipse boundaries $(\pm 2, 0)$ and $(0, \pm 1)$.

4. Shortcut for Linear Modulus Areas

The area bounded by $|ax| + |by| = c$ is a rhombus with area $\frac{2c^2}{|ab|}$. For the region $|x| + |y| = 1$ (which is equivalent to our boundary), the area is $\frac{2(1)^2}{1 \cdot 1} = 2$. Using such shortcuts saves significant time during the JEE Main exam.

5. Common Mistakes to Avoid

Wrong Area Formula: Confusing ellipse area with circle area $\pi r^2$. Always identify 'a' and 'b' correctly.
Symmetry Errors: Forgetting to multiply by 4 when calculating area using only the first quadrant.
Integration Errors: Mistaking $\int \sqrt{a^2 - x^2} dx$. The formula is $\frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a}$.
Overlap Miscalculation: Subtracting the wrong region because the graph wasn't sketched properly.

6. Step-by-Step Graphing Strategy

Identify standard curves (Circle, Ellipse, Parabola, Hyperbola).
Find intersection points by solving equations simultaneously.
Sketch the region to determine which curve is "Upper" and which is "Lower".
Set up the integral: $\int_{x_1}^{x_2} [y_{upper} - y_{lower}] dx$.

7. Vertical vs Horizontal Strips

Sometimes, integrating with respect to $y$ ($x_{right} - x_{left}$) is easier. In this problem, both ways are symmetric, but horizontal strips are often ignored by students. Practicing both ensures versatility.

8. Exam Relevance

Area under curve questions appear in almost every shift of JEE Main. They often combine modulus, greatest integer functions ($[x]$), or trigonometric functions. Mastering the "Area of Ellipse" and "Area of Modulus regions" covers roughly 30% of the possible variations in this chapter.

9. Advanced Perspective (JEE Advanced)

In Advanced, the regions might be bounded by polar coordinates or involve rotation. However, the fundamental principle remains: Area is the accumulation of infinitesimal strips. The use of double integrals ($\iint dA$) is also a valid (though usually longer) method for these problems.

10. Summary Checklist

Standard form of ellipse achieved?
Area of ellipse $\pi ab$ calculated?
Modulus lines plotted?
Boundary intersection points verified?
Subtraction performed correctly?

Frequently Asked Questions (FAQs)

1. What is the area of a standard ellipse? The area of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $\pi ab$.

2. How do you find the area of the region bounded by |x| + |y| = k? The area is always $2k^2$. For $k=1$, the area is $2(1)^2 = 2$.

3. Can we solve this using only integration? Yes, but it is much slower. You would integrate from -2 to 2 for the ellipse and -1 to 1 for the modulus lines.

4. Is the region $y = 1 - |x|$ and $y = |x| - 1$ a square? Yes, it forms a square with vertices at $(1,0), (-1,0), (0,1), (0,-1)$.

5. Why is the answer not 2π + 2? Because the question asks for the area "outside" the modulus region, so we must subtract it from the total area.

6. What if the ellipse was 4x² + y² = 4? Then $a=1$ and $b=2$. The area would still be $2\pi$, but the shape would be vertical.

7. How to identify if a curve is inside another? Check the boundary points. Since all vertices of the square satisfy $x^2 + 4y^2 \le 4$, the square is inside the ellipse.

8. What is the integration of √(4 - x²)? It is $\frac{x}{2}\sqrt{4-x^2} + 2\sin^{-1}\frac{x}{2}$.

9. Does "outside the region" mean infinite area? No, it says "inside the ellipse AND outside the region", which limits the area to the space between the two boundaries.

10. Is this a common JEE Main topic? Yes, Area Under Curves is a high-weightage topic often combined with modulus functions.

Expert Contribution by: JEE NEET Experts

Jee neet experts, 10 year experience | Specialists in Calculus and Coordinate Geometry.