Find the vertices and foci of the hyperbola with equation quantity x plus 4 squared divided by 9 minus the quantity of y minus 5 squared divided by 16 = 1

Vertices at (-7, 5) and (-1, 5).

Foci at (-9, 5) and (1,5).

Step-by-step explanation:

(x + 4)²/9 - (y - 5)²/16 = 1

The standard form for the equation of a hyperbola with centre (h, k) is

(x - h²)/a² - (y - k)²/b² = 1

Your hyperbola opens left/right, because it is of the form x - y.

Comparing terms, we find that

h = -4, k = 5, a = 3, y = 4

In the general equation, the coordinates of the vertices are at (h ± a, k).

Thus, the vertices of your parabola are at (-7, 5) and (-1, 5).

The foci are at a distance c from the centre, with coordinates (h ± c, k), where c² = a² + b².

c² = 9 + 16 = 25, so c = 5.

The coordinates of the foci are (-9, 5) and (1, 5).

The Figure below shows the graph of the hyperbola with its vertices and foci.

No a triangle cannot be formed because when you add the two smallest number together (which would be 11) the have to be bigger than the number it is "against" ( which would be 12)

Your answer should be 1/3. i hope i could ! good luck : )