Question 1121383
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The endpoints of the transverse axis are the vertices.  Since the transverse axis is horizontal, the branches of the hyperbola open right and left.  So the general form of the equation is<br>
{{{(x-h)^2/a^2-(y-k)^2/b^2 = 1}}}<br>
The coordinates of the vertices tell us the center (h,k) is (0,0), so the equation is<br>
{{{x^2/a^2-y^2/b^2 = 1}}}<br>
The slopes of the two asymptotes are b/a and -b/a; since the equation of one of the asymptotes is y=x, that means b=a.<br>
a is the distance from the center to each vertex; so a=4.<br>
Then, since b=a, b=4.<br>
And so the equation is<br>
{{{x^2/4^2-Y^2/4^2 = 1}}}<br>
or<br>
{{{x^2/16-y^2/16 = 1}}}