To find the vertices and foci of the conic section: frac{(x - 4)^{2}}{5^{2}} - frac{(y + 3)^{2}}{6^{2}}=1

To find the vertices and foci of the conic section:
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Concept used: Write the general equation of the hyperbola. $\frac{{\left(x-h\right)}^{2}}{{a}^{2}}-\frac{{\left(y-k\right)}^{2}}{{b}^{2}}=1$ The centre of the hyperbola is (h,k), the vertices are $\left(h±a,k\right)$, and foci are $\left(h±c,k\right)$ Write the expression for c. $c=\sqrt{{a}^{2}+{b}^{2}}$ The given equation is, ${\left(\frac{x-4}{5}\right)}^{2}-{\left(\frac{y+3}{6}\right)}^{2}=1$
$\frac{{\left(x-4\right)}^{2}}{{5}^{2}}-\frac{{\left(y+3\right)}^{2}}{{6}^{2}}=1$ Write the general equation of the hyperbola. $\frac{{\left(x-h\right)}^{2}}{{a}^{2}}-\frac{{\left(y-k\right)}^{2}}{b}=1$ Compare the given equation with the general equation of the hyperbola to get, $h=4$
$k=-3$
$a=5$
$b=6$ Write the expression for c. $c=\sqrt{{a}^{2}+{b}^{2}}$ Substitute 6 for a and 5 for b. $c=\sqrt{{5}^{2}+{6}^{2}}$
$=\sqrt{25+36}$
$=\sqrt{61}$ The vertices of the conic section are, $\left(h±a,k\right)=\left(4±5,-3\right)$
$=\left(4+5,-3\right)\left(4-5,-3\right)$
$=\left(9,-3\right)\left(-1,-3\right)$ The foci of the conic section are, $\left(h±a,k\right)=\left(4±\sqrt{61},-3\right)$
$=\left(4+\sqrt{61},-3\right)\left(4-\sqrt{61},-3\right)$ Therefore, the vertices and foci of the conic section are $\left(9,-3\right)\left(-1,-3\right)$ and $\left(4+\sqrt{61},-3\right)\left(4-\sqrt{61},-3\right)$ respectively.