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Chapter 6: Problem 62

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

Group the like terms and rearrange the given equation

Let's rearrange the equation by grouping the like terms together: \(4y^{2} - 4y - 2x^{2} - 8x -15 = 0\) rearranges to \(4(y^{2} - y) - 2(x^{2} + 4x) - 15 = 0\).
02

Complete the square on both x and y

To complete the square on the y terms, we add and subtract \((1/2)^2\), which is 1/4. To do this on the x terms, we add and subtract \((4/2)^2\), which is 4. This gives us \(4[(y - 0.5)^2 - 0.25] - 2[(x + 2)^2 - 4] - 15 = 0\).
03

Simplify the equation and find standard form

After simplifying, our equation becomes \((y - 0.5)^2 - (x + 2)^2 = 1\). We need to divide all terms by 1, so we get \((y - 0.5)^2 /1 - (x + 2)^2 /1 = 1\). This is in the standard form of a hyperbola, \(\frac{(y-h)^2}{a^2} - \frac{(x-k)^2}{b^2} = 1\). Here, h = 0.5, k = -2 are the coordinates of the center of the hyperbola, and a = 1, b = 1 are the lengths of the semi-major and semi-minor axes respectively.

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