Problem 21 Show that the graph of the equat... [FREE SOLUTION]

Chapter 10: Problem 21

Show that the graph of the equation $$ \frac{x^{2}}{15-c}-\frac{y^{2}}{c-6}=1 $$ is (a) a hyperbola with foci $(\pm 3,0)$ if $615$

Short Answer

Expert verified

The graph is a hyperbola if $6 < c < 15 $, an ellipse if $c < 6$, and doesn't correspond to a real graph if $c > 15$.

Step by step solution

Identify the general form of the equation

Compare the given equation $ \frac{x^{2}}{15-c}-\frac{y^{2}}{c-6}=1 $ to the general forms of conics. The equation resembles the standard form of a hyperbola $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $ when $ a^2 > 0 $.

Conditions for a hyperbola

Identify the conditions under which the given equation represents a hyperbola. For the expression $ 15-c $ as the denominator of $ x^2 $ to be positive, we need $ 15-c > 0 $ or $ c < 15 $. Similarly, $ c-6 > 0 $ or $ c > 6 $ for the other term, leading to the condition $ 6 < c < 15 $.

Check foci of the hyperbola

For a hyperbola $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $, the foci are $ (\pm \sqrt{a^2 + b^2}, 0) $. Here, $ a^2 = 15-c $ and $ b^2 = c-6 $. Set $ \sqrt{a^2 + b^2} = 3 $, leading to $ a^2 + b^2 = 9 $, consistent with $ a^2 = 15-c $ and $ b^2 = c-6 $ when $ 6 < c < 15 $.

Conditions for an ellipse

For an ellipse in the form $ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $, both $ a^2 $ and $ b^2 $ must be positive. The given scenario for an ellipse is $ c < 6 $ where $ c-6 < 0 $, thus flipping the sign of the $ y^2 $ term, aligning with an ellipse.

Case when c > 15

When $ c > 15 $, both $ 15-c < 0 $ and $ c-6 > 0 $, thus making the equation $ -\frac{x^2}{a^2} = \frac{y^2}{b^2} - 1 $ which isn't a standard conic form but instead represents two parallel lines or no real graph because there's no satisfying algebraic form for a conic.

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Key Concepts

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

Hyperbola

A hyperbola is one of the classic conic sections, formed by intersecting a double cone with a plane. The most common form of a hyperbola is written as $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $. This reveals its structure:

The numerator, $ x^2 $, means the hyperbola opens along the x-axis when it is positive.
The signs and values of $ a^2 $ and $ b^2 $ dictate the shape and orientation of the hyperbola.

Hyperbolas have two separate curves, known as branches, mirroring each other around their respective axis. What makes a hyperbola distinct is its foci:

The foci are outside the hyperbola. They are special points equidistant to any point on the hyperbola.
To find the foci, use the formula $ (\pm \sqrt{a^2 + b^2}, 0) $, ensuring that $ a^2 + b^2 $ remains constant.

When examining $ \frac{x^2}{15-c} - \frac{y^2}{c-6} = 1 $ for the condition $ 6 < c < 15 $, the equation indeed forms a hyperbola, where the foci calculate to $ (\pm 3, 0) $, as given by the solution.

Ellipse

An ellipse is an elegant shape that resembles a flattened circle or an elongated oval. The mathematical representation of an ellipse takes the form of $ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $. Characteristics of an ellipse include:

Both $ a^2 $ and $ b^2 $ are positive, which defines the stretch along the x-axis and y-axis respectively.
An ellipse will always be a closed curve, unlike the open branches of a hyperbola.

In an ellipse, the main feature is the two focal points, or foci:

These foci lie within the ellipse and are crucial in defining its shape. The sum of distances from any point on the ellipse to the foci is constant.
While the foci aren't specifically used in the given equation, understanding their importance can help visualize the ellipse's symmetry and balance.

When $ c < 6 $ in $ \frac{x^2}{15-c} - \frac{y^2}{c-6} = 1 $, it reconfigures to resemble this scenario. Here, ensuring that both segments of the equation are positive ensures you're dealing with an ellipse.

Foci

Foci are a central concept not only for hyperbolas and ellipses but all conic sections. They represent the unique fixed points that help define the properties and shape of the conics.

In hyperbolas, the foci are located outside the branches, driving each section to open outward. The distance between foci helps describe the hyperbola's separation and shape.
In ellipses, the foci are internal. They determine the oval shape's specific flattening, with the distance affecting the ellipse's eccentricity.

The coordinates of the foci allow us to better understand and construct these conic sections correctly:

For a hyperbola $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $, the foci are located at $ (\pm \sqrt{a^2 + b^2}, 0) $.
In the case of ellipses $ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $, their location is $ (\pm \sqrt{a^2 - b^2}, 0) $, assuming $ a > b $.

Understanding the role of foci enables complex mathematical concepts to become intuitive, ensuring a solid grasp of the symmetry and priorities of conic sections.

91影视

Show that the graph of the equation $$ \frac{x^{2}}{15-c}-\frac{y^{2}}{c-6}=1 $$ is (a) a hyperbola with foci \((\pm 3,0)\) if \(615\)

Short Answer

Step by step solution

Identify the general form of the equation

Conditions for a hyperbola

Check foci of the hyperbola

Conditions for an ellipse

Case when c > 15

Key Concepts

Hyperbola

Ellipse

Foci

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