Question

Find the vertices and foci of the hyperbola with equation quantity x plus one squared divided by sixteen minus the quantity of y plus five squared divided by nine = 1

Answer 1

Answer:

The vertices are (3 , -5) , (-5 , -5)

The foci are (4 , -5) , (-6 , -5)

Step-by-step explanation:

* Lets study the equation of the hyperbola

- The standard form of the equation of a hyperbola with  

  center (h , k) and transverse axis parallel to the x-axis is

  (x - h)²/a² - (y - k)²/b² = 1

- The length of the transverse axis is 2 a

- The coordinates of the vertices are  (h  ±  a  ,  k)

- The coordinates of the foci are (h ± c , k), where c² = a² + b²

- The distance between the foci is  2c

* Now lets solve the problem

- The equation of the hyperbola is (x + 1)²/16 - (y + 5)²/9 = 1

* From the equation

# a² = 16 ⇒ a = ± 4

# b² = 9 ⇒ b = ± 3

# h = -1

# k = -5

∵ The vertices are (h + a , k) , (h - a , k)

∴ The vertices are (-1 + 4 , -5) , (-1 - 4 , -5)

* The vertices are (3 , -5) , (-5 , -5)

∵ c² = a² + b²

∴ c² = 16 + 9 = 25

∴ c = ± 5

∵ The foci are (h ± c , k)

∴ The foci are (-1 + 5 , -5) , (-1 - 5 , -5)

* The foci are (4 , -5) , (-6 , -5)

Answer 2

Answer:

Vertices: (3,-5) (-5,-5)

Foci: (-6,-5) (4,-5)

Step-by-step explanation:

(x+1)^2/16-(y+5)^2/9 =1

formula: (x-h)^2/a^2 -(y-k)^2/b^2=1

in this case...

a^2=16      b^2=9

h=-1 k=-5

a=4           b=3

v=(h+/-a,k)

v1=(-1+4,-5)=

v1=(3,-5)

v2=(-1-4, -5) =

v2=(-5,-5)

Foci=(h+/-c,k)

F1=(h-c,k)

=(-1-5,-5)

f1=(-6,-5)

F2=(h+c,k)

=(-1+5, -5)

F2=(4,-5)

Hope this helps! :)