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Chapter 16: Problem 200

The parabolas \(y^{2}=4 x\) and \(x^{2}=4 y\) divide the square region bounded by the lines \(x=4, y=4\) and the coordinate axes. If \(S_{1}, S_{2}, S_{3}\) are respectively the areas of these parts numbered from top to bottom; then \(S_{1}: S_{2}\) : \(S_{3}\) is (A) \(1: 2: 1\) (B) \(1: 2: 3\) (C) \(2: 1: 2\) (D) \(1: 1: 1\)

Short Answer

Expert verified
The ratio \(S_1 : S_2 : S_3\) is \(1 : 2 : 1\). (Answer: (A))

Step by step solution

01

Understand the Region

We are given two parabolas, \(y^2 = 4x\) and \(x^2 = 4y\), and a square region bounded by \(x = 4\), \(y = 4\), and the coordinate axes. The task is to find the ratio of areas into which the parabolas divide this square.
02

Sketch the Region

Visualize the square on the positive coordinate plane. It stretches from (0,0) to (4,4). The parabola \(y^2 = 4x\) opens to the right, and \(x^2 = 4y\) opens upwards. Sketch these curves inside the square for clarity.
03

Find Intersection Points

To find the intersection points of the parabolas within the square, set \(y^2 = 4x\) equal to \(x^2 = 4y\). Solving yields the intersection points at (0,0) and (4,4), as these are the only points within the square region that solve both equations simultaneously.
04

Divide the Region

The parabolas divide the square into parts. Since they intersect at (4,4), splitting the problem, they form regions above and below this line within the square, and along the x-axis and y-axis from (0,0) to (4,4).
05

Calculate Area of Each Region

Use integration to find the areas. For \(S_1\), integrate the vertical distance between \(x = \frac{y^2}{4}\) and the line \(y = 4\) from \(y = 0\) to \(y = 4\). For \(S_2\), integrate between the curves horizontally. For \(S_3\), integrate the horizontal distance between \(y = \frac{x^2}{4}\) and \(x = 4\) from \(x = 0\) to \(x = 4\).
06

Perform Calculations

Perform the integrations:- \(S_1 = \int_{0}^{4} (4 - \frac{y^2}{4}) \, dy = \frac{16}{3}\)- \(S_3 = \int_{0}^{4} (4 - \frac{x^2}{4}) \, dx = \frac{16}{3}\)- Subtract these from the total area of the square, 16, to find \(S_2 = 16 - 2\times\frac{16}{3} = \frac{32}{3}\).
07

Determine the Ratio

Calculate the ratio \(S_1 : S_2 : S_3\) by simplifying the fractional areas. The ratio is \(\frac{16}{3} : \frac{32}{3} : \frac{16}{3}\), simplifying to \(1 : 2 : 1\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Coordinate Geometry
Coordinate geometry is a fascinating blend of algebra and geometry. It involves plotting points, lines, and curves on an x-y coordinate system to study their properties.
In this exercise, we have two parabolas defined by the equations \( y^2 = 4x \) and \( x^2 = 4y \). Each parabola represents a distinctive curved line with unique properties:
  • The parabola \( y^2 = 4x \) is symmetric about the x-axis and opens to the right.
  • The parabola \( x^2 = 4y \) is symmetric about the y-axis and opens upward.
These geometric shapes help in dividing a certain area within a defined region, which in this case is a square bounded by lines at \( x=4 \) and \( y=4 \), along with the coordinate axes.
By understanding how these curves interact within the square, we can visualize how they partition it into different regions.
Area Calculation
Area calculation in coordinate geometry often involves finding the region enclosed by curves and lines. For this problem, the total region is the square bounded by the axes and lines at \( x=4 \) and \( y=4 \).
To determine the areas \( S_1, S_2, \) and \( S_3 \) made by the parabolas dividing the square:
  • \( S_1 \) and \( S_3 \) are found using definite integrals between the curves and respective boundaries (either axes or the lines \( x=4 \) and \( y=4 \)).
  • \( S_2 \) is the remaining area after subtracting \( S_1 \) and \( S_3 \) from the total area of the square, which is 16.
These areas reveal how much of the square is occupied by each subdivided region.
Intersection Points
Intersection points are where two or more curves meet. In our scenario, to find where the parabolas intersect inside the square, we need to solve the equations \( y^2 = 4x \) and \( x^2 = 4y \) simultaneously.
When we solve these equations, two points emerge that lie within the bounds of the square:
  • The origin \((0, 0)\) which is common for both curves.
  • The point \((4, 4)\) where the end of one curve meets the boundary of the square.
These points help in dividing the square into smaller, distinct sections that can be explored separately for area calculation.
Integration in Mathematics
Integration is a fundamental tool in calculus used for finding areas under curves. It allows us to calculate exact areas enclosed within curves and straight lines.
In this problem, integration helps in determining the areas \( S_1 \) and \( S_3 \):
  • For \( S_1 \), integrate vertically from \( y=0 \) to \( y=4 \) to find the area between the curve \( y^2 = 4x \) (solved as \( x = \frac{y^2}{4} \)) and the line \( y=4 \).
  • For \( S_3 \), integrate horizontally from \( x=0 \) to \( x=4 \) between the curve \( x^2 = 4y \) (solved as \( y = \frac{x^2}{4} \)) and the line \( x=4 \).
Both integra perform with limits set by the region boundaries and curves help in finding precise values for these areas, offering insight into how each section of the square is covered.

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Most popular questions from this chapter

The intercepts on \(x\)-axis made by tangents to the curve, \(y=\int_{0}^{x}|t| d t, x \in R\), which are parallel to the line \(y=2 x\), are equal to (A) \(\pm 2\) (B) \(\pm 3\) (C) \(\pm 4\) (D) \(\pm 1\)

\(\lim _{n \rightarrow \infty} \frac{\left(1^{2}+2^{2}+3^{2}+\ldots+n^{2}\right)\left(1^{3}+2^{3}+3^{3}+\ldots+n^{3}\right)}{\left(1^{6}+2^{6}+3^{6}+\ldots+n^{6}\right)}\) is equal to (A) \(\frac{7}{12}\) (B) \(\frac{12}{7}\) (C) \(\frac{5}{12}\) (D) None of these

The area included between the curves \(x^{2}+y^{2}=a^{2}\) and \(\sqrt{|x|}+\sqrt{|y|}=\sqrt{a}(a>0)\) is (A) \(\left(\pi+\frac{2}{3}\right) a^{2}\) (B) \(\left(\pi-\frac{2}{3}\right) a^{2}\) (C) \(\frac{2}{3} a^{2}\) (D) \(\frac{2 \pi}{3} a^{2}\)

The value of the integral \(\int_{0}^{\pi} \sqrt{1+4 \sin ^{2} \frac{x}{2}-4 \sin \frac{x}{2}} d x\) equals [2014] (A) \(\pi-4\) (B) \(\frac{2 \pi}{3}-4-4 \sqrt{3}\) (C) \(4 \sqrt{3}-4\) (D) \(4 \sqrt{3}-4-\frac{\pi}{3}\)

The value of \(\int_{-\pi}^{\pi} \frac{\cos ^{2} x}{1+a^{x}} d x, a>0\), is (A) \(a \pi\) (B) \(\frac{\pi}{2}\) (C) \(\frac{\pi}{a}\) (D) \(2 \pi\)
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