Question 1035355
1.The major axis is on the line connecting the foci, which is the y-axis.
The center of the hyperbola is halfway between the foci, at (0,0), the origin,
so the minor axis (perpendicular to the y-axis through the center) is the x-axis.
The equation for a parabola centered at the origin, with the y-axis as its major axis is
{{{y^2/a^2-x^2/b^2=1}}} with {{{a>0}}} and {{{b>0}}} .
The focal distance, {{{c}}} , is half the distance between the foci.
In this case, it is {{{c=(4-(-4))/2=4}}} .
The semimajor axis, {{{a}}} , is the distance between the center and each vertex.
The major axis is the vertical segment connecting vertices (0,-a), (0,a),
which are on the major axis, between the foci, so {{{0<a<4}}} .
The semi-minor axis is {{{b}}} .
The asymptotes cross at the center of the hyperbola (in this case (0,0), the origin),
and have equations {{{y=(a/b)x}}} and {{{-(a/b)x}}} .
So far, we have
{{{drawing(200,400,-3,3,-6,6,grid(0),
triangle(-5,-10,5,-10,0,0),triangle(5,10,-5,10,0,0),
green(rectangle(-1.79,-3.58,1.79,3.58)),
circle(0,-4,0.1),circle(0,4,0.1),
circle(0,-3.58,0.1),circle(0,3.58,0.1),
locate(-1.8,2,green(a)),locate(-1.8,-1.6,green(a)),
locate(1.5,2,green(a)),locate(1.5,-1.6,green(a)),
locate(-1,3.58,green(b)),locate(0.8,3.58,green(b)),
locate(0.8,1.6,c)
)}}}
The values {{{a}}} and {{{b}}} determine a rectangle,
passing through the vertices,
with sides measuring {{{2a}}} and {{{2b}}} ,
and diagonals measuring {{{2c}}} .
The axes and the asymptotes divide that rectangle into 8 right trianlges with legs {{{a}}} and {{{b}}} , and hypotenuse {{{c}}} , so
{{{c^2=a^2+b^2}}} .
Since the asymptotes have equations {{{y=2x}}} and {{{y=-2x}}} ,
in this case {{{a/b=2}}} --> {{{a=2b}}} --> {{{a^2=4b^2}}} ,
and since {{{c=4}}} ,
{{{4^2=4b^2+b^2}}}
{{{16=5b^2}}}
{{{b^2=16/5}}}
{{{a^2=4*16/5=64/5}}} .
So, the equation for the hyperbola in this problem is
{{{y^2/(64/5)-x^2/(16/5)=1}}} ,
or {{{y^2/12.8-x^2/3.2=1}}} ,
or {{{5y^2/64-5x^2/16=1}}} ,
or {{{y^2/64-c^2/16=1/5}}} .