SOLUTION: Problem: 3yy+24y-xx-2x+41=0
have to decide if it is a circle, ellipse, parabola, or hyperbola.
it it is a hyperbola the equation has to be in this form:
(x-h)(x-h)/aa-(y-k)(y-k
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Question 600372: Problem: 3yy+24y-xx-2x+41=0
have to decide if it is a circle, ellipse, parabola, or hyperbola.
it it is a hyperbola the equation has to be in this form:
(x-h)(x-h)/aa-(y-k)(y-k)/bb=1 or (y-k)(y-k)/aa-(x-h)(x-h)/bb=1
Parabola: y=a(x-h)(x-h)+k or x=a(y-k)(y-k)+h
then graph please.
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
Problem: 3yy+24y-xx-2x+41=0
have to decide if it is a circle, ellipse, parabola, or hyperbola.
it it is a hyperbola the equation has to be in this form:
(x-h)(x-h)/aa-(y-k)(y-k)/bb=1 or (y-k)(y-k)/aa-(x-h)(x-h)/bb=1
Parabola: y=a(x-h)(x-h)+k or x=a(y-k)(y-k)+h
then graph please
**
3y^2+24y-x^2-2x+41=0
complete the square
3(y^2+8y+16)-(x^2+2x+1)=-41+48-1
3(y+4)^2-(x+1)^2=6
divide by 6
(y+4)^2/2-(x+1)^2/6=1
This is an equation of a hyperbola with vertical transverse axis (y-term listed first)
Its standard form: (y-k)^2/a^2-(x-h)^2/b^2=1, (h,k)=(x,y) coordinates of center
For given equation:
center: (-1,-4)
a^2=2
a=√2
length of vertical transverse axis=2a=2√2≈2.83
b^2=6
b=√6
length of conjugate axis=2b=2√6≈4.9
see graph below
y=(2+(x+1)^2/3)^.5-4
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