# What are the solutions to the system of conics?

Drag the ordered pairs into the box to correctly complete the table.

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• 4 weeks ago

From  (1),  y^2 = 2x^2 + 4

So using also  (2),

2x^2 + 4 = 8x + 8 ,

2x^2 - 8x - 4 = 0 ,

x^2 - 4x        = 2 ,

x^2 - 4x + 4 = 2 + 4 ,

(x + 2)^2 = 6 ,

x = - 2 ± √6 .

Then

y^2 = 8 (-1 ± √6)

\  y = ± 2√2 √(-1 ± √6) , four values   ...   ?

But  -1 - √6 < 0 , so cannot be a value of  y^2 .

Thus, the solutions points are

(-2 + √6 . ± 2√2 √(-1 + √6) ) .

• 4 weeks ago

y^2 / 4 - x^2 / 2 = 1

y^2 - 2x^2 = 4

y^2 = 8 * (x + 1)

8 * (x + 1) - 2x^2 = 4

4 * (x + 1) - x^2 = 2

4x + 4 - x^2 = 2

-x^2 + 4x + 2 = 0

x^2 - 4x - 2 = 0

x^2 - 4x = 2

x^2 - 4x + 4 = 2 + 4

(x - 2)^2 = 6

x - 2 = +/- sqrt(6)

x = 2 +/- sqrt(6)

When choosing between which function to substitute x into, always choose the one with the lower degree.  The higher degree formulas will give extraneous answers

y^2 = 8 * (x + 1)

y^2 = 8 * (2 +/- sqrt(6) + 1)

y^2 = 8 * (3 +/- sqrt(6))

y^2 = 4 * (6 +/- 2 * sqrt(6))

y = +/- 2 * sqrt(6 +/- 2 * sqrt(6))

(2 + sqrt(6) , 2 * sqrt(6 + 2 * sqrt(6)))

(2 + sqrt(6) , -2 * sqrt(6 + 2 * sqrt(6)))

(2 - sqrt(6) , 2 * sqrt(6 - 2 * sqrt(6)))

(2 - sqrt(6) , -2 * sqrt(6 - 2 * sqrt(6)))