Math  /  Geometry

Questionection 8.2 Homework - Hyperbolas Question 8, 8.2.79 HW Score: 41.67%,541.67 \%, 5 of 12 points Part 1 of 3 Points: 0 of 1
Write the equation in standard form for a hyperbola. Find the center and the vertices of the hyperbola. 16y29x296y72x+144=016 y^{2}-9 x^{2}-96 y-72 x+144=0
Choose the correct standard form below. A. (y3)29(x+4)216=1\frac{(y-3)^{2}}{9}-\frac{(x+4)^{2}}{16}=1 B. (x+4)216(y3)29=1\frac{(x+4)^{2}}{16}-\frac{(y-3)^{2}}{9}=1 C. (x+4)29(y3)216=1\frac{(x+4)^{2}}{9}-\frac{(y-3)^{2}}{16}=1 D. (y3)216(x+4)29=1\frac{(y-3)^{2}}{16}-\frac{(x+4)^{2}}{9}=1

Studdy Solution

STEP 1

1. The given equation is that of a hyperbola.
2. We need to rewrite the equation in standard form.
3. Identify the center and vertices of the hyperbola.

STEP 2

1. Rearrange the equation to group xx and yy terms.
2. Complete the square for both xx and yy terms.
3. Rewrite the equation in standard form.
4. Identify the center and vertices of the hyperbola.

STEP 3

Rearrange the equation to group xx and yy terms:
16y296y9x272x+144=0 16y^2 - 96y - 9x^2 - 72x + 144 = 0
Group the yy terms and the xx terms:
(16y296y)(9x2+72x)=144 (16y^2 - 96y) - (9x^2 + 72x) = -144

STEP 4

Complete the square for the yy terms:
Factor out 16 from the yy terms:
16(y26y) 16(y^2 - 6y)
Complete the square:
y26y(y3)29 y^2 - 6y \rightarrow (y - 3)^2 - 9
Thus:
16((y3)29)=16(y3)2144 16((y - 3)^2 - 9) = 16(y - 3)^2 - 144
Complete the square for the xx terms:
Factor out -9 from the xx terms:
9(x2+8x) -9(x^2 + 8x)
Complete the square:
x2+8x(x+4)216 x^2 + 8x \rightarrow (x + 4)^2 - 16
Thus:
9((x+4)216)=9(x+4)2+144 -9((x + 4)^2 - 16) = -9(x + 4)^2 + 144

STEP 5

Substitute the completed squares back into the equation:
16(y3)21449(x+4)2+144=144 16(y - 3)^2 - 144 - 9(x + 4)^2 + 144 = -144
Simplify:
16(y3)29(x+4)2=144 16(y - 3)^2 - 9(x + 4)^2 = -144
Divide the entire equation by 144-144 to get it into standard form:
(y3)29(x+4)216=1 \frac{(y - 3)^2}{9} - \frac{(x + 4)^2}{16} = 1

STEP 6

Identify the center and vertices of the hyperbola:
The standard form of the hyperbola is:
(y3)29(x+4)216=1 \frac{(y - 3)^2}{9} - \frac{(x + 4)^2}{16} = 1
The center of the hyperbola is (h,k)=(4,3)(h, k) = (-4, 3).
The vertices are determined by the yy-term, since it is positive and comes first:
a2=9a=3 a^2 = 9 \Rightarrow a = 3
Vertices are at (h,k±a)=(4,3±3)(h, k \pm a) = (-4, 3 \pm 3), which are (4,6)(-4, 6) and (4,0)(-4, 0).
The correct standard form is:
(y3)29(x+4)216=1 \boxed{\frac{(y - 3)^2}{9} - \frac{(x + 4)^2}{16} = 1}
The center is (4,3)(-4, 3) and the vertices are (4,6)(-4, 6) and (4,0)(-4, 0).

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