Need second intersection point of two parabolas.

Question

Y=8x^2 and x=8y^2

clearly (0,0) is one point of intersection. They should have another at I believe (1/8,1/8). How is that derived?

TIA

Answer 1

Equation (1):  y = 8x²

Equation (2):  x = 8y²

 

Substitute 8y² for x (from equation 2) in equation (1):

 

y = 8x²

y = 8(8y²)²

y = 512y⁴

1/512 = y³

 

1/8 = y

 

Now plug y = 1/8 into the second equation (2):

 

x = 8y²

x= 8(1/8)²

 

x = 1/8

Answer 2

Hi Judith,

 

We have two equations and two unknowns, so let's try to solve it like other such problems. That is, let's take x=8y², and solve for y, and substitute into into y=8x².

 

Solving x=8y² for y, we get y = ±sqrt(x/8). Note that wee must include the "+" and "-" roots.

 

Substituting this into y=8x², we get ±sqrt(x/8)=8x². We can square each side and get rid of the +/-; the result is (x/8)=64x⁴. Taking all terms to one side of the "=", we obtain 0=64x⁴-(x/8). We factor out the "x" to obtain 0=x*[64x³-(1/8)]. Thus, there are two solutions, x=0 (which you identified right away) and then

 

    64x³ - 1/8 = 0, or multiplying by 8, 8*64x³ - 1 = 0, or 512x³ = 1, or x³ = 1/512

 

I think I'll let you finish it from here and see if you get (x=1/8, y=1/8).

 

Good luck ... YW!