Determine two pairs of polar coordinates for the point (5, 5) with 0° ≤ θ < 360°.

a. (5 square root 2, 225°), (-5 square root 2, 45°)

b. (5 square root 2, 315°), (-5 square root 2, 135°)

c. (5 square root 2, 135°), (-5 square root 2, 315°)

d. (5 square root 2, 45°), (-5 square root 2, 225°)

find an equation in standard form for the hyperbola with vertices at (0, ±9) and foci at (0, ±10).

a. y squared over 81 minus x squared over 100 = 1
b. y squared over 81 minus x squared over 19 = 1
c. y squared over 19 minus x squared over 81 = 1
d. y squared over 100 minus x squared over 81 = 1

First part

The answer is (5 square root 2, 45°), (-5 square root 2, 225°) ⇒ answer (d)

Second part

The equation in standard form for the hyperbola is y²/81 - x²/19 = 1 ⇒ answer(b)

Step-by-step explanation:

First part:

* Lets study the Polar form and the Cartesian form

- The important difference between Cartesian coordinates and

  polar coordinates:

# In Cartesian coordinates there is exactly one set of coordinates

  for any given point.

# In polar coordinates there is an infinite number of coordinates

   for a given point. For instance, the following four points are all

   coordinates for the same point.

# In the polar the coordinates the origin is called the pole, and

  the x axis is called the polar axis.

# The angle measurement θ can be expressed in radians

   or degrees.

- To convert from Cartesian Coordinates (x , y) to

  Polar Coordinates (r , θ)

# r = ± √(x² + y²)

# θ = tan^-1 (y / x)

* Lets solve the problem

- The point in the Cartesian coordinates is (5 , 5)

∵ x = 5 and y = 5

∴ r = ± √(5² + 5²) = ± √50 = ± 5√2

∴ tanФ = (5/5) = 1

∵ tanФ is positive

∴ Angle Ф could be in the first or third quadrant

∵ Ф = tan^-1 (1) = 45°

∴ Ф in the first quadrant is 45°

∴ Ф in the third quadrant is 180 + 45 = 225°

* The answer is (5√2 , 45°) , (-5√2 , 225°)

Second part:

* Lets study the standard form of the hyperbola equation

- The standard form of the equation of a hyperbola with  

  center (0 , 0) and transverse axis parallel to the y-axis is

  y²/a² - x²/b² = 1, where

•the length of the transverse axis is 2a

•the coordinates of the vertices are (0 , ±a)

•the length of the conjugate axis is 2b

•the coordinates of the co-vertices are (±b , 0)

•      the coordinates of the foci are (0 , ± c),  

•the distance between the foci is 2c, where c² = a² + b²

* Lets solve our problem

∵ The vertices are (0 , 9) and (0 , -9)

∴ a = ± 9 ⇒ a² = 81

∵ The foci at (0 , 10) , (0 , -10)

∴ c = ± 10

∵ c² = a² + b²

∴ (10)² = (9)² + b² ⇒ 100 = 81 + b² ⇒ subtract 81 from both sides

∴ b² = 19

∵ The equation is  y²/a² - x²/b² = 1

∴  y²/81 - x²/19 = 1

* The equation in standard form for the hyperbola is y²/81 - x²/19 = 1