Question #171144

A. Given the following information about the hyperbola, find both its standard and general form:

1. FOCI: (-4, 0) (4, 0); VERTICES: (-3, 0) (3, 0)

2. FOCI: (-3, -3) (-3, 13); VERTICES: (-3, 0) (-3, 10)


B. Given the standard form of the equation of hyperbola, find the following, then sketch the graph:


A. orientation

B. opening of branches

C. center

D. transverse axis

E. conjugate axis

F. distance of the foci

G. vertices

H. foci

I. co-vertices

J. asymptotes


1.(x+4)^2/9 − (y+3)^2/4 = 1


2.(y+2)^2/36 − (x−1)^2/49 = 1


C. Convert each general form to standard form of the equation of hyperbola: (10 points)

1. 9y^2 − 4x^2 − 18y + 24x − 63 = 0


2. 9x^2 - 4y^2 - 90x + 32y - 163 = 0





1
Expert's answer
2021-03-18T10:06:17-0400

A. 1. Basic “horizontal” hyperbola:

Equation: x2a2y2b2=1\dfrac{x^2}{a^2 }-\dfrac{y^2 }{b^2 }=1

c2=a2+b2c^2=a^2+b^2

Foci: (c,0),(c,0)(-c, 0), (c, 0)

Vertices: (a,0),(a,0)(-a, 0), (a, 0)

a=3,c=4a=3, c=4


b2=4232=7b^2=4^2 -3^2=7

The standard form of an equation of a hyperbola


x29y27=1\dfrac{x^2}{9 }-\dfrac{y^2 }{7 }=1

The general form of an equation of a hyperbola


6x29y263=06x^2-9y^2-63=0

2. The center is (h,k)(h, k) and the vertices are (h,k±b)(h, k\pm b)

Equation: (yk)2b2(xh)2a2=1\dfrac{(y-k)^2}{b^2 }-\dfrac{(x-h)^2 }{a^2 }=1

Foci: (h,kc),(h,k+c)(h, k-c), (h, k+c)

h=3,b=5,c=8,k=5h=-3, b=5, c=8, k=5


a2=8252=39a^2=8^2-5^2=39

The standard form of an equation of a hyperbola


(y5)225(x+3)239=1\dfrac{(y-5)^2}{25}-\dfrac{(x+3)^2}{39 }=1

The general form of an equation of a hyperbola


25x2150x+39y2390y225=0-25x^2-150x+39y^2-390y-225=0

B.

1.


(x+4)29(y+3)24=1\dfrac{(x+4)^2}{9}-\dfrac{(y+3)^2}{4 }=1

A. Horizontal oriented hyperbola 

B. Opens left and right.

C. Center (4,3).(-4,-3).

D. Major (transverse) axis length: 6

E. Minor (conjugate) axis length: 4

F. c=13c=\sqrt{13}

Focal Parameter: 41313\dfrac{4\sqrt{13}}{13}

G. Vertices: (7,3),(1,3)(-7, -3), (-1, -3)

H. Foci: (413,3),(4+13,3)(-4-\sqrt{13}, -3), (-4+\sqrt{13}, -3)

I. Co-vertices: (4,5),(4,1)(-4, -5), (-4, -1)

J. Asymptotes:y=23x173,y=23x13y=-\dfrac{2}{3}x-\dfrac{17}{3}, y=\dfrac{2}{3}x-\dfrac{1}{3}


2.


(y+2)236(x1)249=1\dfrac{(y+2)^2}{36}-\dfrac{(x-1)^2}{49 }=1

A. Verticaloriented hyperbola 

B. Opens upward and downward.

C. Center (1,2).(1,-2).

D. Major (transverse) axis length: 12

E. Minor (conjugate) axis length: 14

F. c=85c=\sqrt{85}

Focal Parameter: 498585\dfrac{49\sqrt{85}}{85}

G. Vertices: (1,8),(1,4)(1, -8), (1, 4)

H. Foci: (1,285),(1,2+85)(1,-2-\sqrt{85}), (1,-2+\sqrt{85})

I. Co-vertices: (6,2),(8,2)(-6, -2), (8, -2)

J. Asymptotes:y=67x87,y=67x207y=-\dfrac{6}{7}x-\dfrac{8}{7}, y=\dfrac{6}{7}x-\dfrac{20}{7}


C.

1.


9y24x218y+24x63=09y^2-4x^2-18y+24x-63=0

9(y22y+1)4(x26x+9)=369(y^2-2y+1)-4(x^2-6x+9)=36

(y1)24(x3)29=1\dfrac{(y-1)^2}{4}-\dfrac{(x-3)^2}{9}=1

2.


9x24y290x+32y163=09x^2-4y^2-90x+32y-163=0

9(x210x+25)4(y28y+16)=3249(x^2-10x+25)-4(y^2-8y+16)=324

(x5)236(y4)281=1\dfrac{(x-5)^2}{36}-\dfrac{(y-4)^2}{81}=1



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