FINDING VERTICES FOCI AND ASYMPTOTES FOR HYPERBOLA

Equation of Hyperbola

Equation of hyperbola which is symmetric about x-axis.

Equation of hyperbola which is symmetric about y-axis.

Use the vertices and asymptotes to graph each hyperbola. Locate the foci and find the equation of asymptotes.

Problem 1 :

Solution :

From the given equation, x² is the first term. Then the hyperbola is symmetric about x-axis.

a² = 9 and b² = 16

a = 3 and b = 4

Center :

C (0, 0)

Vertices :

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

A(3, 0) and A'(-3, 0)

For hyperbola,

c² = a² + b²

c² = 3² + 4²

c² = 9 + 16

c² = 25

c = 5

Foci :

F₁ (c, 0) and F₂ (-c, 0)

F₁ (5, 0) and F₂ (-5, 0)

Equation of asymptotes :

Since the hyperbola is symmetric about x-axis, the equations of asymptotes are y = (-b/a)x and y = (b/a)x

Here a = 3 and b = 4

y = (-4/3) and y = (4/3) x

Problem 2 :

Solution :

From the given equation, x² is the first term. Then the hyperbola is symmetric about x-axis.

a² = 100 and b² = 64

a = 10 and b = 8

Center :

C (0, 0)

Vertices :

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

A(10, 0) and A'(-10, 0)

For hyperbola,

c² = a² + b²

c² = 10² + 8²

c² = 100  + 64

c² = 164

c = √164

= 2√41

Foci :

F₁ (c, 0) and F₂ (-c, 0)

F₁ (2√41, 0) and F₂ (-2√41, 0)

Equation of asymptotes :

Since the hyperbola is symmetric about x-axis, the equations of asymptotes are y = (-b/a)x and y = (b/a)x

Here a = 10 and b = 8

y = (-8/10) and y = (8/10) x

y = (-4/5) and y = (4/5) x

Problem 3 :

Solution :

From the given equation, y² is the first term. Then the hyperbola is symmetric about y-axis.

a² = 16 and b² = 36

a = 4 and b = 6

Center :

C (0, 0)

Vertices :

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

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

For hyperbola,

c² = a² + b²

c² = 4² + 6²

c² = 16 + 36

c² = 52

c = √52

= 2√13

Foci :

F₁ (0, c) and F₂ (0, -c)

F₁ (0, 2√13) and F₂ (0, 2√13)

Equation of asymptotes :

Since the hyperbola is symmetric about y-axis, the equations of asymptotes are y = (-a/b)x and y = (a/b)x

Here a = 4 and b = 6

y = (-4/6)x and y = (4/6) x

y = (-2/3)x  and y = (2/3) x

Problem 4 :

4y² - x² = 1

Solution :

4y² - x² = 1

From the given equation, y² is the first term. Then the hyperbola is symmetric about y-axis.

a² = 1/4 and b² = 1

a = 1/2 and b = 1

Center :

C (0, 0)

Vertices :

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

A(0, 1/2) and A'(0, -1/2)

For hyperbola,

c² = a² + b²

c² = (1/2)² + 1²

c² = (1/4) + 1

c² = 5/4

c = √5/2

Foci :

F₁ (0, c) and F₂ (0, -c)

F₁ (0, √5/2) and F₂ (0, -√5/2)

Equation of asymptotes :

Since the hyperbola is symmetric about y-axis, the equations of asymptotes are y = (-a/b)x and y = (a/b)x

Here a = 1/2 and b = 1

y = (-1/2)x and y = (1/2) x

Problem 5 :

(x - 3)² - 4(y + 3)² = 4

Solution :

(x - 3)² - 4(y + 3)² = 4

Dividing by 4 on both sides.

From the given equation, x² is the first term. Then the hyperbola is symmetric about x-axis.

a² = 4 and b² = 1

a = 2 and b = 1

Center :

 Comparing the given equation with

 (x - h)²/a² - (y - k)²/b² = 1

C (h, k) ==> (3, -3)

Vertices :

A(h + a, k) and A'(h - a, k)

h = 3, a = 2

h + a => 3 + 2

h - a => 3 - 2 = 1

A(5, -3)  and A'(1, -3)

For hyperbola,

c² = a² + b²

c² = 2² + 1²

c² = 4 + 1

c² = 5

c = √5

Foci :

F₁(h + c, k) and F₂ (h - c, k)

Here h = 3 and c = √5

h + c ==> 3 + √5 

h - c ==> 3 - √5 

F₁ (3 + √5 , -3) and F₂ (3 - √5 , -3)

Equation of asymptotes :

Since the hyperbola is symmetric about x-axis, the equations of asymptotes are 

y - k = ±(b/a)(x - h)

y - (-3) = ±(1/2)(x - 3)

y + 3 = ±(1/2)(x - 3)

Problem 6 :

Solution :

From the given equation, y² is the first term. Then the hyperbola is symmetric about y-axis.

a² = 4 and b² = 16

a = 2 and b = 4

Center :

 Comparing the given equation with

 (y - k)²/a² - (x - h)²/b² = 1

C (h, k) ==> (1, -2)

Vertices :

A(h, k + a) and A'(h, k - a)

h = 1, a = 2

k + a => -2 + 2 = 0

k - a => -2 - 2 = -4

A(1, 0)  and A'(1, -4)

For hyperbola,

c² = a² + b²

c² = 2² + 4²

c² = 4 + 16

c² = 20

c = 2√5

Foci :

F₁(h, k + c) and F₂ (h, k - c)

Here h = 1 and c = 2√5

k + c ==> -2 + 2√5 

k - c ==> -2 - 2√5 

F₁ (1, -2 + 2√5 ) and F₂ (1, -2 - 2√5)

Equation of asymptotes :

Since the hyperbola is symmetric about y-axis, the equations of asymptotes are 

y - k = ±(a/b)(x - h)

a = 2, b = 4, h = 1 and k = -2

y - (-2) = ±(2/4)(x - 1)

y + 2 = ±(2/4)(x - 1)

y + 2 = ±(1/2)(x - 1)