PROBLEMS ON CONIC SECTIONS

Problem 1 :

A hyperbolic mirror can be used to take panoramic photos, if the camera is pointed toward the mirror with the lens at one focus of the hyperbola. Write the equation of the hyperbola that can be used to model a mirror that has a vertex 5 inches from the center of the hyperbola and a focus 1 inch in front of the surface of the mirror. Assume the mirror has a horizontal transverse axis and the hyperbola centered at (0, 0).

Solution :

From the given information,

Distance between center and vertex = 5

a = 5

Distance between center and focus = 6

ae = 6

5e = 6

e = 6/5

b² = a² (e² - 1)

Applying the values of a and e, we get

b² = 5² ((6/5)² - 1)

= 25 ((36/25) - 1)

= 25 ( (36 - 25) / 25) )

b² = 9

Since it has horizontal transverse axis, the hyperbola is symmetric about x-axis.

x²/a² - y²/b² = 1

x²/25 - y²/9 = 1

Problem 2 :

Skips designs tracks for amusement park rides. For a new design, the track will be elliptical. If the ellipse is placed on a large coordinate grid with center at (0, 0). The equation

models the path of the track . The units are given in yards. How long is the major axis of the track.? Explain how you find the distance.

Solution :

a² = 8100

b² = 3600

Since the denominator of y² is greater the elliptical shape is symmetric about y-axis.

Major axis is y-axis. Length of major axis = 2a

From a² = 8100, a = 90

From b² = 3600, b = 60

Length of major axis = 2(90)= 180 units.

Problem 3 :

The equation of parabola is

24y = (x - 2)² - 48

identify the vertex, focus and directrix of the parabola.

Solution :

24y = (x - 2)² - 48

y = ((x - 2)² - 48)/24

4a = 1/24

a = 1/4(24)

a = 1/96

h = 2 and k = -2

From the given equation, we know that the parabola is symmetric about y-axis and open upward since a is  > 0

Center : (h, k) ==> (2, -2)

Focus : (h, a) ==> (2, 1/96)

Equation of directrix : y = -a

y = -1/96

Problem 4 :

The hyperbola has vertices (±4, 0) and focus (5, 0). What is the standard form equation of the hyperbola ?

Solution :

Vertices (±4, 0) and focus (5, 0)

A(4, 0) and A'(-4, 0)

Center of the hyperbola = midpoint of the vertices

= (4 + (-4)) /2 , (0 + 0)/2

= (0, 0)

From the given information, we know that the hyperbola is symmetric about x-axis.

a = 4,

Distance between center and focus = c = 5

c² = a² + b²

5² = 4² + b²

b² = 25 - 16

b² = 9

Problem 5 :

State the vertices, foci, and asymptotes of the hyperbola with the equation 

20x² - 25y² = 100

Solution :

20x² - 25y² = 100

Dividing by 100, we get

x²/5 - y²/4 = 1

The hyperbola is symmetric about x-axis.

a² = 5 and b² = 4

a = √5 and b = 2

Center :

(0, 0)

Vertices :

A (a, 0) and A'(-a, 0)

A (√5, 0) and A' (-√5, 0)

Foci :

F₁(c, 0) and F₂ (-c, 0)

c² = a² + b²

c² = 5 + 4

c² = 9

c = 3

F₁(3, 0) and F₂ (-3, 0)

Asymptotes :

x = ±(b/a) 

x = ±(2/√5) 

x = (2/√5)  and x = -(2/√5)

Problem 6 :

Find the equation that models the path of a satellite if its path is a hyperbola, a = 45000 km and c = 71000. Assume that the center of the hyperbola is the origin and the transverse axis is horizontal.

Solution :

The hyperbola is symmetric about x-axis with the center at origin.

a = 45000 km and c = 71000

c² = a² + b²

(71000)²= (45000)² + b²

5041000000 -2025000000 = b²

b² = 3016000000

So, the required equation is 

(x² / 2025000000) - (y² / 3016000000)= 1