/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 30 Determine whether the following ... [FREE SOLUTION] | 91Ó°ÊÓ

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Determine whether the following equations describe a parabola, an ellipse, or a hyperbola, and then sketch a graph of the curve. For each parabola, specify the location of the focus and the equation of the directrix; for each ellipse, label the coordinates of the vertices and foci, and find the lengths of the major and minor axes; for each hyperbola, label the coordinates of the vertices and foci, and find the equations of the asymptotes. $$10 x^{2}-7 y^{2}=140$$

Short Answer

Expert verified
Answer: The given equation represents a hyperbola with vertices at (±√14, 0), foci at (±√34, 0), and asymptotes given by ±(√20/√14)y = x.

Step by step solution

01

Rewrite the given equation in a recognizable form

We are given the equation: $$10x^2 - 7y^2 = 140.$$ First, we start by dividing both sides by 140 to simplify the equation: $$\frac{10x^2}{140} - \frac{7y^2}{140} = 1$$ Which simplifies to: $$\frac{x^2}{14} - \frac{y^2}{20} = 1.$$
02

Determine the type of curve

Observe that the equation has a positive term containing \(x^2\) and a negative term containing \(y^2\). This means our equation represents a hyperbola.
03

Find the vertices and foci

For a hyperbola in the form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), the vertices are located at \((\pm a, 0)\). In our equation, \(a^2 = 14\), so \(a = \sqrt{14}\). Therefore, the vertices are at \((\sqrt{14}, 0)\) and \((-\sqrt{14}, 0)\). The distance from the center to each focus is given by \(c = \sqrt{a^2 + b^2}\). In our equation, \(b^2 = 20\), so \(c = \sqrt{14 + 20} = \sqrt{34}\). Hence, the foci are located at \((\sqrt{34}, 0)\) and \((-\sqrt{34}, 0)\).
04

Find the equations of the asymptotes

For a hyperbola in the form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), the equations of the asymptotes are \(\pm\frac{b}{a}y = x\). For our equation, we get the asymptotes: $$\pm\frac{\sqrt{20}}{\sqrt{14}}y = x.$$
05

Final answer

The equation represents a hyperbola with vertices at \((\sqrt{14}, 0)\) and \((-\sqrt{14}, 0)\), foci at \((\sqrt{34}, 0)\) and \((-\sqrt{34}, 0)\), and asymptotes given by \(\pm\frac{\sqrt{20}}{\sqrt{14}}y = x\).

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Key Concepts

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

Conic Sections
The term conic sections refers to the curves that are obtained by intersecting a plane with a double-napped cone. These curves can be identified as parabolas, ellipses, and hyperbolas depending on the angle at which the plane intersects the cone.

For example, when the plane intersects the cone parallel to its side, a parabola is formed. If the plane cuts the cone at a sharp angle but does not go parallel to any side, an ellipse is produced. However, if the plane intersects both naps of the cone, a hyperbola results. These geometric figures are fundamental in various areas of mathematics and science, such as astronomy, physics, and engineering.

The equation given in the exercise, after simplification, represents a hyperbola - a conic section where the plane's intersection angle is steeper than the cone's side, but not perpendicular to the cone's axis, which yields the characteristic two separate curves.
Vertices and Foci of Hyperbola
Understanding the vertices and foci of a hyperbola is critical when analyzing its shape and properties. For a hyperbola of the form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), the vertices lie along the x-axis at points \(\pm a, 0\). These points mark the 'tips' of the hyperbola, where it is closest to being 'pointed'.

The foci of a hyperbola, on the other hand, are points from which distances to any point on the hyperbola have a constant difference. These foci are found along the axis of symmetry of the hyperbola, which in this case is the x-axis. The distance to each focus from the center is given by \(c = \sqrt{a^2 + b^2}\), which reflects how 'spread out' the hyperbola is.

In the context of the exercise, we calculated the positions for both the vertices, \(\pm\sqrt{14}, 0\), and the foci, \(\pm\sqrt{34}, 0\), offering a clear illustration of these concepts.
Equations of Asymptotes
The equations of the asymptotes for a hyperbola provide the slopes of the lines that the hyperbola approaches but never touches—these are the visual 'guidelines' for the open ends of the hyperbola. For the standard form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), the equations of the asymptotes are determined by \(\frac{b}{a}\), which results in the lines \(\pm\frac{b}{a}y = x\).

In the given exercise, by finding the values of 'a' and 'b' from the hyperbola equation, we can state the asymptote equations as \(\pm\frac{\sqrt{20}}{\sqrt{14}}y = x\). These lines graphically represent the paths the arms of the hyperbola tend toward both in the positive and negative directions indefinitely, thereby outlining the spread of the hyperbola's branches.

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