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Chapter 6: Problem 46

Find the standard form of the equation of the hyperbola with the given characteristics. Vertices: \((3,0),(3,4)\) asymptotes: \(y=\frac{2}{3} x, y=4-\frac{2}{3} x\)

Short Answer

Expert verified
The standard form of the equation of the hyperbola with the given characteristics is \((y-2)^2/4 - (x-3)^2/9 = 1\).

Step by step solution

01

Finding the center

The center of the hyperbola is the midpoint of the line segment joining the vertices. Applying the midpoint formula, the center \(h, k\) is given by \((h, k) = \left( \frac{(x_1+x_2)/2}, (y_1+y_2/2) \right) = \left( \frac{(3+3)/2}, (0+4)/2 \right) = (3,2).\)
02

Finding 'a' value

The distance from the center to either vertex is 'a'. It is calculated as \(a = |y_2 - k| = |4 - 2| = 2\).
03

Finding 'b' value

The 'b' value is related to the slope of the asymptotes, such that \(a/b\) equals to the slope of the asymptotes. Given the slope of the asymptotes is \(2/3\), we can therefore find b as \(b = a/\text{slope of the asymptotes} = 2/(2/3) = 3\).
04

Writing the equation of the hyperbola

Substituting all the values back into the standard form equation of the hyperbola, we have \((y-k)^2/a^2 - (x-h)^2/b^2 = 1\), therefore, our equation becomes \((y-2)^2/4 - (x-3)^2/9 = 1\).

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Most popular questions from this chapter

(a) Show that the distance between the points \(\left(r_{1}, \theta_{1}\right)\) and \(\left(r_{2}, \theta_{2}\right)\) is \(\sqrt{r_{1}^{2}+r_{2}^{2}-2 r_{1} r_{2} \cos \left(\theta_{1}-\theta_{2}\right)}\) (b) Simplify the Distance Formula for \(\theta_{1}=\theta_{2} .\) Is the simplification what you expected? Explain. (c) Simplify the Distance Formula for \(\theta_{1}-\theta_{2}=90^{\circ}\) Is the simplification what you expected? Explain.

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