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Chapter 6: Problem 54

Classify the graph of the equation as a circle, a parabola, an ellipse, or a hyperbola. $$4 y^{2}-2 x^{2}-4 y-8 x-15=0$$

Short Answer

Expert verified
The given equation represents a hyperbola.

Step by step solution

01

Arrange the Equation

Rearrange the given equation so that variables are grouped together, starting with the squares: \(4y^2 - 4y - 2x^2 -8x - 15 = 0\). Further, divide everything by -2 to simplify: \(2x^2 + 4x -2y^2 + 2y + 7.5 = 0\).
02

Complete the Square for x and y Terms

Rewrite the equation in the form of [a(x-h)^2] and [b(y-k)^2]. For the x-terms \(2x^2+4x\) can be written as \(2(x^2+2x)\), the number to complete the square (called 'c') is \(\frac{2^2}{4} = 1\), and thus \(2(x^2 + 2x + 1 - 1)\). The same method should be applied to the y-terms, \(2y^2-2y\) becomes \(-2(y^2-y)\), 'c' will be \(\frac{1^2}{4} = 0.25\), and thus \(-2(y^2 - y + 0.25 - 0.25)\).
03

Simplify the Equation

The equation simplifies to \(2(x+1)^2 - 2(y-0.5)^2 + 2 - 7.5 = 0\), which further simplifies to \(2(x+1)^2 - 2(y-0.5)^2 = 5.5\). Divide by 5.5 to ensure that the equation equals 1: \(\frac{(x+1)^2}{2.75} - \frac{(y-0.5)^2}{5.5} = 1\).
04

Classify the Equation

A comparison of this equation with the standard forms of the equation for a circle \(\frac{(x-h)^2}{r^2} + \frac{(y-k)^2}{r^2} = 1\), an ellipse \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\), a hyperbola \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\) and a parabola \(y = a(x - h)^2 + k\) shows that this equation is a hyperbola since the equation fits the standard form of a hyperbola.

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Most popular questions from this chapter

In Exercises 79-82, find the exact value of the trigonometric function given that \(u\) and \(v\) are in Quadrant IV and \(\sin u=-\frac{3}{5}\) and \(\cos v=1 / \sqrt{2}\). $$ \cos (u+v) $$

\(r=4+3 \cos \theta\)

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In Exercises 73-78, solve the trigonometric equation. $$ 2 \cot x=5 \cos \frac{\pi}{2} $$
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