SOLUTION: What point is the intersection of the graphs of y = 3x, x^2 + 4y^2 = 37, and y^2 - x^2 = 8 ?

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Question 475404: What point is the intersection of the graphs of y = 3x, x^2 + 4y^2 = 37, and y^2 - x^2 = 8 ?

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 + 4y^2 = 37
x^2/37 + y^2/9.25 = 1 Ellipse C(0,0) and radius = 5 (See below)
y^2 - x^2 = 8
y^2/8 - x^2/8 = 1 Hyperbola opening up and down (See below)
y = 3x Line: Pt(0,0) and Pt(1,3) on the Line
algebraically: substituting 3x for y
x^2 + 36x^2 = 37 x = ± 1 and y = ± 3
Graphs: ellipse, hyperbola and Line intersect at:(1,3) and (-1,-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 )