SOLUTION: What point is the intersection of the graphs of x2 + y2 = 34 and x2 -4y = 13?

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Question 475401: What point is the intersection of the graphs of x2 + y2 = 34 and x2 -4y = 13?

Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 
Hi,
What point is the intersection of the graphs:
x^2 + y^2 = 34 circle C(0,0) and radius = sqrt(34)
x^2 -4y = 13 OR y = .25x^2 - 13/4 Parabola opening upward V(0,-3.25)
algebraically: substituting (13+4y) for x^2
13+4y + y^2 = 34
y^2 + 4y -21 = (y + 7)(y-3) |tossing out y = -7 as Extraneous
y = 3 then x^2 = 25 , x = ± 5
Graphs: parbola and circle intersect at:( 5,3) and (-5,3)


Standard Form of an Equation of a Circle is %28x-h%29%5E2+%2B+%28y-k%29%5E2+=+r%5E2
where Pt(h,k) is the center and r is the radius
Standard Form of an Equation of an Ellipse is %28x-h%29%5E2%2Fa%5E2+%2B+%28y-k%29%5E2%2Fb%5E2+=+1+
where Pt(h,k) is the center and a and b are the respective vertices distances from center.
Standard Form of an Equation of an Hyperbola opening right and left is:
%28x-h%29%5E2%2Fa%5E2+-+%28y-k%29%5E2%2Fb%5E2+=+1 where Pt(h,k) is a center with vertices 'a' units right and left of center.
Standard Form of an Equation of an Hyperbola opening up and down is:
%28y-k%29%5E2%2Fb%5E2+-+%28x-h%29%5E2%2Fa%5E2+=+1 where Pt(h,k) is a center with vertices 'b' units up and down from center.
The vertex form of a parabola opening up or down, y=a%28x-h%29%5E2+%2Bk where(h,k) is the vertex.
The standard form is %28x+-h%29%5E2+=+4p%28y+-k%29, where the focus is (h,k + p)
The vertex form of a parabola opening right or left, x=a%28y-k%29%5E2+%2Bh where(h,k) is the vertex.
The standard form is %28y+-k%29%5E2+=+4p%28x+-h%29, where the focus is (h +p,k )