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Question 474233: What is the focus of the parabola with the equation (x 1)2 + 32 = 8y?
What is the directrix of the parabola with the equation y + 3 = 1/10(x + 2)2?
What is the axis of symmetry of the parabola with the equation x 4 = 1/4(y + 1)2?
Find the slopes of the asymptotes of a hyperbola with the equation y2 = 36 + 4x2.
Thank you for your time its much appreciated
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! What is the focus of the parabola with the equation (x 1)2 + 32 = 8y?
What is the directrix of the parabola with the equation y + 3 = 1/10(x + 2)2?
What is the axis of symmetry of the parabola with the equation x 4 = 1/4(y + 1)2?
Find the slopes of the asymptotes of a hyperbola with the equation y2 = 36 + 4x2.
Thank you for your time its much appreciated
**
What is the focus of the parabola with the equation (x 1)2 + 32 = 8y?
(x1)^2+32=8y
divide by 8
(1/8)(x-1)^2+4=y
(1/8)(x-1)^2=(y-4)
(x-1)^2=8(y-4)
This is a parabola with a vertical axis of symmetry of the standard form: (x-h)^2=4p(y-k), (h,k) being the (x,y) coordinates of the vertex. Parabola opens upward.
For given equation:
vertex: (1,4)
axis of symmetry: x=1
4p=8
p=2
focus:(1,6) (2 units above the vertex on the axis of symmetry, y=1
..
What is the directrix of the parabola with the equation y + 3 = 1/10(x + 2)2?
y+3=(1/10)(x+2)^2
multiply by 10
10(y+3)=(x+2)^2
(x+2)^2=10(y+3)
This is a parabola with a vertical axis of symmetry, same standard form as previous problem.
vertex: (-2,-3)
axis of symmetry: x=-2
4p=10
p=10/4=5/2
directrix: a line, y=11/2 (5/2 units below the vertex perpendicular to the axis of symmetry, x=-2)
..
What is the axis of symmetry of the parabola with the equation x 4 = 1/4(y + 1)2?
x4=(1/4)(y+1)^2
multiply by 4
(y+1)^2=4(x-4)
This is a parabola with a horizontal axis of symmetry of the standard form: (y-k)^2=4p(x-h), (h,k) being the (x,y) coordinates of the vertex. Parabola opens rightwards
For given equation:
vertex: (4,-1)
axis of symmetry: y=-1
..
Find the slopes of the asymptotes of a hyperbola with the equation y2 = 36 + 4x^2
y^2=36+4x^2
y^2-4x^2=36
divide by 36
y^2/36-x^2/9=1
This is a hyperbola with a vertical transverse axis of the standard form: (y-k)^2/a^2-(x-h)^2/b^2=1,
(h,k) being the (x,y) coordinates of the center.
For given problem:
center: (0,0)
a^2=36
a=6
b^2=9
b=3
slope of asymptotes=ħa/b=ħ6/3=ħ2
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