Question

System of conics help

Answer 1

Answer:

Solutions are (6,3) and (-2,3).

Step-by-step explanation:

From the first equation given we can write:

$(x-2)^{2} = 16-(y-3)^{2}$  

substituting for $(x-2)^{2}$ in the second equation given we get

$\frac{16-(y-3)^{2}}{16} + \frac{(y-3)^{2}}{64} = 1$

$\frac{-(y-3)^{2}}{16} + \frac{(y-3)^{2}}{64} =0$

∴ y=3

Putting y=3 in the first equation we get

$(x-2)^{2} = 16$

x-2 = ±4

Hence x=6 or x=-2.