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Find the intersection points, x^2+y^2-16y+39=0, y^2-x^2-9=0

User NSZombie
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8.1k points

1 Answer

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x^2 + y^2 - 16y + 39 = 0
-x^2 + y^2 - 9 = 0
-----------------------------------
2y^2 - 16y + 30 = 0 iff
y^2 - 8y + 15 = 0 iff
(y - 3)(y - 5) = 0 iff
y = 3 or y = 5.

Now plug in y value into any of the given equations (the second one is easier) to find respective x values.

y^2 - x^2 - 9 = 0
(3)^2 - x^2 - 9 = 0 iff
9 - x^2 - 9 = 0 iff
x^2 = 0 iff
x = 0
(0, 3)

y^2 - x^2 - 9 = 0
(5)^2 - x^2 - 9 = 0 iff
25 - x^2 - 9 = 0 iff
16 - x^2 = 0 iff
(4 - x)(4 + x) = 0 iff
x = 4 or x = -4
(4, 5), (-4, 5)

We have three points of intersection, which are (0, 3), (4, 5), and (-4, 5).
User Jatha
by
8.4k points

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