Question 256402
{{{(x + 3)^2-4*(y-2)^2 = 4}}}
{{{(x + 3)^2/4-(y-2)^2 = 1}}}
{{{a=2}}}
{{{b=1}}}
{{{h=-3}}}
{{{k=2}}}
The vertices are then located at,
({{{h+a}}},{{{k}}}) and ({{{h-a}}},{{{k}}})
({{{-3+2}}},{{{2}}}) and ({{{-3-2}}},{{{2}}})
({{{-1}}},{{{2}}}) and ({{{-5}}},{{{2}}})
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.
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{{{ graph( 300, 300, -10, 4, -4, 10, 2+sqrt(((x+3)^2-4)/4), 2-sqrt(((x+3)^2-4)/4)) }}}