SOLUTION: Find the points where these conics intersect. 16𝑥^2 − 5𝑦^2 = 64 and 16𝑥^2 + 25𝑦^2 − 96𝑥 = 256

Algebra ->  Trigonometry-basics -> SOLUTION: Find the points where these conics intersect. 16𝑥^2 − 5𝑦^2 = 64 and 16𝑥^2 + 25𝑦^2 − 96𝑥 = 256      Log On


   

Question 911572: Find the points where these conics intersect.
16𝑥^2 − 5𝑦^2 = 64 and 16𝑥^2 + 25𝑦^2 − 96𝑥 = 256
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
1.16x%5E2-5y%5E2=64
2.16x%5E2+%2B+25y%5E2-96x+=+256
From eq. 1,
5y%5E2=16x%5E2-64
25y%5E2=80x%5E2-320
Substitute into eq. 2,
16x%5E2+%2B+80x%5E2-320-+96x+=+256
96x%5E2-96x-320=256
96x%5E2-96x-576=0
x%5E2-x-6=0
%28x-3%29%28x%2B2%29=0
Two solutions in x:
x-3=0
x=3
Then,
5y%5E2=16%289%29-64
5y%5E2=144-64
5y%5E2=80
y%5E2=16
y=4 and y=-4
.
.
.
x%2B2=0
x=-2
Then,
5y%5E2=16%284%29-64
5y%5E2=0
y=0
So the points are,
(3,4), (3,-4), and (-2,0)