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

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