Question 911572
1.{{{16x^2-5y^2=64}}}
2.{{{16x^2 + 25y^2-96x = 256}}}
From eq. 1,
{{{5y^2=16x^2-64}}}
{{{25y^2=80x^2-320}}}
Substitute into eq. 2,
{{{16x^2 + 80x^2-320- 96x = 256}}}
{{{96x^2-96x-320=256}}}
{{{96x^2-96x-576=0}}}
{{{x^2-x-6=0}}}
{{{(x-3)(x+2)=0}}}
Two solutions in x:
{{{x-3=0}}}
{{{x=3}}}
Then,
{{{5y^2=16(9)-64}}}
{{{5y^2=144-64}}}
{{{5y^2=80}}}
{{{y^2=16}}}
{{{y=4}}} and {{{y=-4}}}
.
.
.
{{{x+2=0}}}
{{{x=-2}}}
Then,
{{{5y^2=16(4)-64}}}
{{{5y^2=0}}}
{{{y=0}}}
So the points are,
(3,4), (3,-4), and (-2,0)