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Question 44360:
1.What is the equation of the directrix of the parabola with equation x=4y2+16y+19?
2. What is the length of the major axis of the ellipse with equation 4(x+4)2+9(y-1)2+36?
3. What are the foci of the ellipse with equation x2+4y2=36?
4. What are the slpoes of the asymptotes of the hyperbola with equation 4x2-y2+8x-6y=9?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 1.What is the equation of the directrix of the parabola with equation x=4y2+16y+19?
x-19+16=4(y^2+4y+4)
x-3=4(y+2)^2
(y+2)^2=(1/4)(x-3)
Vertex is at (3,-2); 4p=1/4 so p=1/16;
Parabola opens to the right so the directrix is 1/16 to the left of the
vertex. Directrix is x=3-(1/16)=47/16
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2. What is the length of the major axis of the ellipse with equation 4(x+4)^2+9(y-1)^2=36?
Dividing through by 36 get;
[(x+4)^2/9] + [(y-1)^2/4] = 1
a=3 so the length of the major axis is 2a=6
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3. What are the foci of the ellipse with equation x^2+4y^2=36?
Divide through by 36 to get;
[x^2/36] +[y^2/9] = 1
a=6, b=3
c^2=a^2-b^2
c=sqrt(27)=3sqrt3
Center is at (0,0)
So Foci are at (c,0), (-c,0)
Foci: (3sqrt3,0) (-3sqrt3,0)
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4. What are the slopes of the asymptotes of the hyperbola with equation 4x2^-y2+8x-6y=9?
4x^2+8x-(y^2+6y)=9
4(x^2+2x+1)-(y^2+6y+9)=9+4-9
(x+1)^2-[(y+3)^2]/4 =1
Center is at (h,k)=(-1,-3)
a=1;b=2
c^2=a^2+b^2=5; so c=sqrt5
Asymptote forms are y-k=(b/a)(x-h) and y-k=-(b/a)(x-h)
I'll let you make the proper substitutions.
Cheers,
Stan H.
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