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Math  /  Algebra

QuestionFind the center, the vertices, the foci, and the asymptotes. Then draw the graph. (x−2)29−(y+5)216=1\frac{(x-2)^{2}}{9}-\frac{(y+5)^{2}}{16}=1

Studdy Solution

STEP 1

1. The given equation is in the standard form of a hyperbola.
2. The standard form of a hyperbola with a horizontal transverse axis is (x−h)2a2−(y−k)2b2=1\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1.
3. The center of the hyperbola is (h,k)(h, k).
4. The vertices are located aa units from the center along the transverse axis.
5. The foci are located cc units from the center along the transverse axis, where c2=a2+b2c^2 = a^2 + b^2.
6. The asymptotes of a hyperbola with a horizontal transverse axis are given by the equations y=k±ba(x−h)y = k \pm \frac{b}{a}(x-h).

STEP 2

1. Identify the center of the hyperbola.
2. Determine the vertices of the hyperbola.
3. Calculate the foci of the hyperbola.
4. Find the equations of the asymptotes.
5. Sketch the graph of the hyperbola.

STEP 3

Identify the center of the hyperbola by comparing the given equation to the standard form (x−h)2a2−(y−k)2b2=1\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1.
The given equation is: (x−2)29−(y+5)216=1 \frac{(x-2)^2}{9} - \frac{(y+5)^2}{16} = 1
From this, we can see: h=2 h = 2 and k=−5 k = -5 .
Thus, the center of the hyperbola is (2,−5)(2, -5).

STEP 4

Determine the vertices of the hyperbola. The vertices are located aa units from the center along the transverse axis. Here, a2=9a^2 = 9, so a=3a = 3.
Since the transverse axis is horizontal, the vertices are: (h±a,k)=(2±3,−5) (h \pm a, k) = (2 \pm 3, -5)
Calculating these, we find the vertices are: (5,−5)and(−1,−5) (5, -5) \quad \text{and} \quad (-1, -5)

STEP 5

Calculate the foci of the hyperbola. The foci are located cc units from the center along the transverse axis, where c2=a2+b2c^2 = a^2 + b^2.
Given a2=9a^2 = 9 and b2=16b^2 = 16, calculate c2c^2: c2=9+16=25 c^2 = 9 + 16 = 25 c=25=5 c = \sqrt{25} = 5
Thus, the foci are: (h±c,k)=(2±5,−5) (h \pm c, k) = (2 \pm 5, -5)
Calculating these, we find the foci are: (7,−5)and(−3,−5) (7, -5) \quad \text{and} \quad (-3, -5)

STEP 6

Find the equations of the asymptotes. For a hyperbola with a horizontal transverse axis, the asymptotes are given by: y=k±ba(x−h) y = k \pm \frac{b}{a}(x-h)
Substitute b=4b = 4, a=3a = 3, h=2h = 2, and k=−5k = -5 into the formula: y=−5±43(x−2) y = -5 \pm \frac{4}{3}(x-2)
This results in two equations: y=−5+43(x−2)andy=−5−43(x−2) y = -5 + \frac{4}{3}(x-2) \quad \text{and} \quad y = -5 - \frac{4}{3}(x-2)

STEP 7

Sketch the graph of the hyperbola using the center, vertices, foci, and asymptotes.
1. Plot the center at (2,−5)(2, -5).
2. Plot the vertices at (5,−5)(5, -5) and (−1,−5)(-1, -5).
3. Plot the foci at (7,−5)(7, -5) and (−3,−5)(-3, -5).
4. Draw the asymptotes using the equations y=−5+43(x−2)y = -5 + \frac{4}{3}(x-2) and y=−5−43(x−2)y = -5 - \frac{4}{3}(x-2).
5. Sketch the hyperbola opening left and right, approaching the asymptotes as it extends.
The center is (2,−5)(2, -5), the vertices are (5,−5)(5, -5) and (−1,−5)(-1, -5), the foci are (7,−5)(7, -5) and (−3,−5)(-3, -5), and the asymptotes are y=−5+43(x−2)y = -5 + \frac{4}{3}(x-2) and y=−5−43(x−2)y = -5 - \frac{4}{3}(x-2).

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