SOLUTION: Can't get these two - need to choose correct conic from the equation: 1. y^2 - 16x^2 - 14y + 17 = 0 - can't get that one 2. x^2 + y^2 - 20x

SOLUTION: Can't get these two - need to choose correct conic from the equation: 1. y^2 - 16x^2 - 14y + 17 = 0 - can't get that one 2. x^2 + y^2 - 20x - 16y + 64 =0 I said hyper

Question 100047: Can't get these two - need to choose correct conic from the equation:
1. y^2 - 16x^2 - 14y + 17 = 0 - can't get that one
2. x^2 + y^2 - 20x - 16y + 64 =0 I said hyperbola?????
Thanks

Answer by robolthuis(1) (Show Source): You can put this solution on YOUR website!
question 2.
rewrite the formule in the form:
x^2 - 20x + y^2 -16y = -64
next
x^2 -20x is the same as x^2 -20x +100 -100
x^2 -20x +100 is the same as (x-10)^2
SO: x^2 -20x is the same as (x-10)^2 -100
In the same way:
y^2 -16y is the same as (y-8)^2 -64
The complete formule becomes:
(x-10)^2 + (y-8)^2 -164 = -64
or:
(x-10)^2 + (y-8)^2 = 100
this is a circle with centre (10, 8) and radius 10

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