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Chapter 11: Problem 30

Find the equation of the ellipse satisfying the given conditions. Write the answer both in standard form and in the form \(A x^{2}+B y^{2}=C\). Endpoints of the major axis \((卤10,0);\) endpoints of the minor axes \((0,卤4)\)

Short Answer

Expert verified
The equation in standard form is \(\frac{x^2}{100} + \frac{y^2}{16} = 1\) and in general form is \(16x^2 + 100y^2 = 1600\).

Step by step solution

01

Identify the Center of the Ellipse

Since the endpoints of the major axis are \((卤10,0)\), the center of the ellipse is located at the midpoint of these two points. The midpoint is \((0,0)\). Hence, the center of the ellipse is \((0,0)\).
02

Determine Length of the Semi-Major Axis

The distance from the center to the endpoints of the major axis \((卤10,0)\) is 10. Thus, the length of the semi-major axis, \(a\), is \(10\).
03

Determine Length of the Semi-Minor Axis

The endpoints of the minor axis are given as \((0,卤4)\). The distance from the center to these endpoints is 4. Hence, the length of the semi-minor axis, \(b\), is \(4\).
04

Write the Standard Form Equation of the Ellipse

In the standard form equation of the ellipse, for a horizontal major axis, it is given by: \[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1.\] Substituting the values \(a = 10\) and \(b = 4\) gives: \[\frac{x^2}{100} + \frac{y^2}{16} = 1.\]
05

Convert Standard Form to General Form

To convert \(\frac{x^2}{100} + \frac{y^2}{16} = 1\) to the general form \(Ax^2 + By^2=C\), multiply through by 1600 (the least common multiple of 100 and 16): \[ (16)x^2 + (100)y^2 = 1600.\] The general form is \(16x^2 + 100y^2 = 1600\).

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Key Concepts

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

Precalculus
Precalculus is a course that bridges the gap between algebra and calculus. It covers various topics including functions, complex numbers, and trigonometry. The study of precalculus also involves understanding shapes and figures such as ellipses. This is important as it helps you gain insights into their orientation, size, and position on the coordinate plane. Understanding ellipses requires knowledge of graphing equations and transformations, topics commonly discussed in precalculus. By grasping precalculus concepts, you prepare yourself for the more complex functions and calculations of calculus.
Standard Form
The standard form of the ellipse's equation is a crucial starting point when graphing or analyzing an ellipse. It is expressed as:
  • For horizontal major axis: \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \)
  • For vertical major axis: \( \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 \)
This representation makes it simple to determine the orientation and lengths of the axes.
The form \( \frac{x^2}{100} + \frac{y^2}{16} = 1 \) indicates a horizontal major axis, with "100" and "16" corresponding to the squares of the semi-major and semi-minor axes lengths, respectively. The standard form is helpful for plotting an ellipse accurately.
General Form
The general form of an ellipse's equation can be derived from the standard form and is helpful in identifying properties of an ellipse more algebraically. It has the structure \( Ax^2 + By^2 = C \), where \( A \), \( B \), and \( C \) are constants. To convert from standard to general form, each term is cleared of fractions typically by multiplying through by a common factor.
In this exercise, converting \( \frac{x^2}{100} + \frac{y^2}{16} = 1 \) to \( 16x^2 + 100y^2 = 1600 \) involves multiplying by 1600, the least common multiple of 100 and 16. The general form provides a more simplified way to work with and manipulate ellipse equations.
Semi-Major Axis
The semi-major axis of an ellipse is half of the longest diameter, running from the center to the perimeter on the major axis. In simpler terms, it's the radius of the longest line that passes through the center. For the given problem, endpoints of the major axis were \((卤10,0)\). The length of the semi-major axis, \(a\), is calculated as 10, since it is the distance from the center \((0,0)\) to either endpoint of the major axis.
Understanding the semi-major axis helps in visualizing the full stretch of the ellipse's longest direction, which aids in sketching and computing its properties.
Semi-Minor Axis
The semi-minor axis is the radius of the shortest diameter of an ellipse, running perpendicular to the semi-major axis. It measures from the center to the perimeter in the shortest direction. From the problem's context, with minor axis endpoints at \((0,卤4)\), the length of the semi-minor axis, \(b\), is 4. The calculation is straightforward as it is the distance from the center \((0,0)\) to one endpoint of the minor axis.
Knowing the semi-minor axis's length is vital for understanding the ellipse's overall shape and size, especially when drafting or analyzing its geometric features.

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Most popular questions from this chapter

Graph the equations. $$3 x^{2}-2 x y+3 y^{2}+2=0$$

As is the case for most asteroids in our solar system, the orbit of the asteroid Gaspra is located between the orbits of Mars and Jupiter. Given that its distances from the Sun at perihelion and aphelion are \(2.132 \mathrm{AU}\) and \(2.288 \mathrm{AU}\) respectively, compute the length of the semimajor axis of the orbit and the eccentricity. Round both answers to two decimal places. Remark: Within the past decade the Galileo spacecraft (on a mission to Jupiter) took images of Gaspra, which can be viewed on the World Wide Web. For instance, at the time of this writing, one NASA website containing these images is httpi//www.jpl.nasa.gov/galileo. From this web page, use the search feature there to find "Gaspra."

Let \(P\left(x_{1}, y_{1}\right)\) be a point on the ellipse \(\left(x^{2} / a^{2}\right)+\left(y^{2} / b^{2}\right)=1\) Let \(N\) be the point where the normal through \(P\) meets the \(x\) -axis, and let \(F\) be the focus \((-c, 0) .\) Show that \(F N / F P=e,\) where \(e\) denotes the eccentricity.

Let \(\left(x_{1}, y_{1}\right)\) be any point on the ellipse \(\left(x^{2} / a^{2}\right)+\left(y^{2} / b^{2}\right)=1(a>b)\) other than one of the endpoints of the major or minor axis. Show that the normal at \(\left(x_{1}, y_{1}\right)\) does not pass through the origin. Hint: Find the \(y\)-intercept of the normal.

Consider the equation \(\left(x^{2} / 3^{2}\right)+\left(y^{2} / 2^{2}\right)=1\) (a) Use the symmetry tests from Section 1.7 to explain why the graph of this equation must be symmetric about both the \(x\) -axis and the \(y\) -axis. (b) Show that solving the equation for \(y\) yields \(y=\pm \frac{1}{3} \sqrt{36-4 x^{2}}\) (c) Let \(y=\frac{1}{3} \sqrt{36-4 x^{2}} .\) Use the techniques of Section 2.4 to find the domain of this function. (You need to solve the inequality \(36-4 x^{2} \geq 0 .\) ) You should find that the domain is the closed interval [-3,3] (d) Use a calculator to complete the following table. Then plot the resulting points and connect them with a smooth curve. This gives you a sketch of the first quadrant portion of the graph of \(\left(x^{2} / 3^{2}\right)+\left(y^{2} / 2^{2}\right)=1\) $$\begin{array}{llllllll}x & 0 & 0.5 & 1 & 1.5 & 2 & 2.5 & 3 \\\\\hline y=\frac{1}{3} \sqrt{36-4 x^{2}} & 2 & & & & & & 0 \\\\\hline\end{array}$$ (e) Use your graph in part (d) along with the symmetry results in part (a) to sketch a complete graph of \(\left(x^{2} / 3^{2}\right)+\left(y^{2} / 2^{2}\right)=1\).
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