Question

PLEASE HELP!!! WILL GIVE BRANLIEST

Answer 1

If you plot those points on a coordinate plane, you see that your hyperbola is vertical in nature, meaning it will go up from one vertex and down from the other, as opposed to side to side.  The equation for a vertical hyperbola is $\frac{(y-k) ^{2} }{ a^{2} } - \frac{(x-h) ^{2} }{ b^{2} } =1$.  When we look at the graph of the points we plotted as the vertices and the foci, the point (0,0) is in the dead center, which is the center.  So we have our h and k.  From the formula we see need to find a, b, and c.  a is the distance from the center to the vertex, so a = 4, and c is the distance from the center to the focus, so c = 5.  Use that in Pythagorean's Theorem to find b.  $(4) ^{2} + b^{2} = (5)^{2}$ and $16+ b^{2} =25$ and b = 3.  So now we have all the info we need to write the equation: $\frac{(y-0) ^{2} }{16} - \frac{(x-0) ^{2} }{9} =1$.  Simplifying you get $\frac{y ^{2} }{16} - \frac{ x^{2} }{9} =1$.  These are hard and require lots of practice!

Answer 2

Check the picture below, so the hyperbola looks more or less like that.

based on the provided vertices, its center is at the origin, as you see there, the "a" component or traverse axis is 4, the "c" distance is 5.

$\bf \textit{hyperbolas, vertical traverse axis } \\\\ \cfrac{(y- k)^2}{ a^2}-\cfrac{(x- h)^2}{ b^2}=1 \qquad \begin{cases} center\ ( h, k)\\ vertices\ ( h, k\pm a)\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ \qquad \sqrt{ a ^2 + b ^2} \end{cases}\\\\ -------------------------------$

$\bf \begin{cases} h=0\\ k=0\\ c=5\\ a=4 \end{cases}\implies \cfrac{(y-0)^2}{4^2}-\cfrac{(x-0)^2}{b^2}=1 \\\\\\ \stackrel{c}{5}=\sqrt{4^2+b^2}\implies 5^2=16+b^2\implies 25=16+b^2 \\\\\\ 9=b^2\implies \sqrt{9}=b\implies \boxed{3=b} \\\\\\ \cfrac{(y-0)^2}{4^2}-\cfrac{(x-0)^2}{3^2}=1\implies \cfrac{y^2}{16}-\cfrac{x^2}{9}=1$