GRAPH THE ELLIPSE AND IDENTIFY THE CENTER VERTICES AND FOCI

Graph the ellipse and identify the center, vertices and foci.

Problem 1 :

Solution:

b > a

The above ellipse is symmetric about y-axis.

Center:

(0, 0)

Vertices:

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

A(0, 5) and A'(0, -5)

Foci:

c² = a² - b²

c² = 5² - 3²

= 25 - 9

c² = 16

c = √16

c = ±4

F₁(0, 4) F₂(0, -4)

Problem 2 :

4x² + 9y² = 36

Solution:

4x² + 9y² = 36

a² = 9 and b² = 4

a = 3 and b = 2

Since, a > b, the ellipse symmetric about x-axis.

Center:

(0, 0)

Vertices:

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

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

Foci:

c² = a² - b²

c² = 3² - 2²

= 9 - 4

c² = 5

c = ±√5

F₁(√5, 0) F₂(-√5, 0)

Problem 3 :

x² + 4y² = 36

Solution:

x² + 4y² = 36

a² = 36 and b² = 9

a = 6 and b = 3

Since, a > b, the ellipse symmetric about x-axis.

Center:

(0, 0)

Vertices:

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

A(6, 0) and A'(-6, 0)

Foci:

c² = a² - b²

c² = 6² - 3²

= 36 - 9

c² = 27

c = ±3√3

F₁(3√3, 0) F₂(-3√3, 0)

Problem 4 :

Solution:    

Let X = x + 1 and Y = y - 3

Since, b > a, the ellipse symmetric about y-axis.

a² = 4 and b² = 1

a = 2 and b = 1

Center:

(0, 0)

X = 0 and Y = 0

Substitute X = x + 1 and Y = y - 3

x + 1 = 0 and y - 3 = 0

x = -1 and y = 3

The center is (-1, 3).

Vertices:

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

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

The vertices are (-1, 5) and (-1, 1).

Foci:

c² = a² - b²

= 4 - 1

c² = 3

c = √3

Here (h, k) = center of the ellipse

h = -1

k = 3

vertical ellipse = (h, k ± c)

= (-1, 3 ± √3)

Problem 5 :

Solution:

Let X = x - 1 and Y = y + 2

Since, b > a, the ellipse symmetric about y-axis.

a² = 9 and b² = 4

a = 3 and b = 2

Center:

(0, 0)

X = 0 and Y = 0

Substitute X = x - 1 and Y = y + 2

x - 1 = 0 and y + 2 = 0

x = 1 and y = -2

The center is (1, -2).

Vertices:

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

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

The vertices are (1, 1) and (1, -5).

Foci:

c² = a² - b²

= 9 - 4

c² = 5

c = √5

Here (h, k) = center of the ellipse

h = 1

k = -2

vertical ellipse = (h, k ± c)

= (1, -2 ± √5)

Problem 6 :

36(x + 4)² + (y + 3)² = 36

Solution:

36(x + 4)² + (y + 3)² = 36

Since, b > a, the ellipse symmetric about y-axis.

a² = 36 and b² = 1

a = 6 and b = 1

Center:

(0, 0)

X = 0 and Y = 0

Substitute X = x + 4 and Y = y + 3

x + 4 = 0 and y + 3 = 0

x = -4 and y = -3

The center is (-4, -3).

Vertices:

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

A(0, 6) and A'(0, -6)

The vertices are (-4, 3) and (-4, -9).

Foci:

c² = a² - b²

= 36 - 1

c² = 35

c = √35

Here (h, k) = center of the ellipse

h = -4

k = -3

vertical ellipse = (h, k ± c)

= (-4, -3 ± √35)