Question

what are the coordinates of the vertices of the conic section shown below?

(y+2)^2 / 25 - (x-3)^2 / 4 = 1

i just know this will be a hyperbola

Answer 1

the fraction with the "y" is positive, so for this hyperbola is a "vertical traverse axis"

so the vertices will be at \(\bf \cfrac{(y-{\color{blue}{ k}})^2}{{\color{brown}{ a}}^2}-\cfrac{(x-{\color{blue}{ h}})^2}{b^2}=1\qquad vertices\to ({\color{blue}{ h,k}}\pm {\color{brown}{ a}})\)

Answer 2

thank you n_n @jdoe0001