SOLUTION: write the parabola in standard form and graph x^2-4y-28=0 write the ellipse in standard form and graph 2x^2+y^2-4x-4y+4=0 write the hyperbola in standard form and graph x^2-

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: write the parabola in standard form and graph x^2-4y-28=0 write the ellipse in standard form and graph 2x^2+y^2-4x-4y+4=0 write the hyperbola in standard form and graph x^2-      Log On


   

Question 754965: write the parabola in standard form and graph x^2-4y-28=0
write the ellipse in standard form and graph 2x^2+y^2-4x-4y+4=0
write the hyperbola in standard form and graph x^2-8x+11=-y^2
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
write the parabola in standard form and graph:
x^2-4y-28=0
4y=x^2-28
y=(1/4)x^2-7
standard form: y=A(x-h)^2+k, (h,k)=(x,y) coordinates of vertex.
see graph below:
+graph%28+300%2C+300%2C+-10%2C+10%2C+-10%2C+10%2C++x%5E2%2F4-7%29+
..
write the ellipse in standard form and graph
2x^2+y^2-4x-4y+4=0
2x^2-4x+y^2-4y+4=0
complete the square:
2(x^2-2x+1)+(y^2-4y+4)=-4+2+4
2(x-1)^2+(y-2)^2=2
(x-1)^2+(y-2)^2/2=1
see graph below:
y=(2-2(x-1)^2)^.5+2

..
write the hyperbola in standard form and graph
x^2-8x+11=-y^2
x^2-8x+y^2=-11
complete the square:
(x^2-8x+16)+y^2=-11+16
(x-4)^2+y^2=5
(x-4)^2+y^2=5
This is an equation of a circle, not a hyperbola.
see graph below:
y=(5-(x-4)^2)^.5