FINDING CENTER FOCI VERTICES AND DIRECTRIX OF ELLIPSE AND HYPERBOLA

Examples 1-2 : Find center, foci, vertices, and equations of directrices of of the following ellipses :

Example 1 :

Solution :

The given ellipse is symmetric about x-axis.

Center : (0, 0)

Foci :

F(ae, 0)  and  F'(-ae, 0)

Foci are F(4, 0) and F'(-4, 0).

Vertices :

A(a, 0)  and  A'(-a, 0)

A(5, 0)  and  A'(-5, 0) 

Equations of directrices :

x = a/e  and  x = -a/e

Example 2 :

Solution :

The given ellipse is symmetric about y-axis.

Center : (0, 0)

Foci :

F₁ (ae, 0) F₂ (-ae, 0)

Foci are F(0, √7) and F'(0, √7).

Vertices :

A(0, a)  and  A'(0, -a)

A(0, √10)  and  A'(0, -√10) 

Equations of directrices :

y = a/e  and  y = -a/e

Example 3 :

Solution :

The given hyperbola is symmetric about x-axis.

Center : (0, 0)

Foci :

F(ae, 0)  and  F'(-ae, 0)

Foci are F(13, 0) and F'(-13, 0).

Vertices :

A (a, 0) A' (-a, 0) 

A (5, 0) A' (-5, 0) 

Equation of directrices :

x = a/e  and  x = -a/e

Example 4 :

Solution :

The given hyperbola is symmetric about y-axis.

Center : (0, 0)

Foci :

F(0, ae)  and  F'(0, -ae)

Foci are F(0, 5) and F'(0, -5).

Vertices :

A(0, a)  and  A'(0, -a) 

A(0, 4)  and  A'(0, -4) 

Equations of directrices :

y = a/e  and  y = -a/e

Find the eccentricity, centre, foci, vertices of the following ellipses

Example 5 :

(x + 3)²/6 + (y - 5)²/4 = 1

Solution :

(x + 3)²/6 + (y - 5)²/4 = 1

Let X = x + 3 and Y = y - 5

X²/6 + Y²/4 = 1

The ellipse is symmeteric about x-axis.

a² = 6 and b² = 4

a = √6 and b = 2

Center :

Vertices :

Foci :

F(ae, 0)  and  F'(-ae, 0)

e = √1 - (b²/a²)

e = √1 - (4/6)

= √(6 - 4)/6

= √2/6

= √(1/3)

e = 1/√3

ae = √6 (1/√3)

= √2√3(1/√3)

= √2

x + 3 = √2 and x + 3 = -√2

x = √2 - 3 and x = -√2 - 3

F (√2 - 3, 5) F' (-√2 - 3, 5)

Find the eccentricity, centre, foci, vertices of the following ellipses

Example 6 :

36x² + 4y² - 72x + 32y - 44 = 0

Solution :

36x² + 4y² - 72x + 32y - 44 = 0

36x² - 72x + 4y² + 32y - 44 = 0

36(x² - 2x) + 4(y² + 8y) - 44 = 0

36(x² - 2x(1) + 1² - 1²) + 4(y² + 2y(4) + 4² - 4² ) - 44 = 0

36[(x - 1)² - 1] + 4[(y + 4)² - 16] - 44 = 0

36(x - 1)² - 36 + 4(y + 4)² - 64 - 44 = 0

36(x - 1)² + 4(y + 4)² - 36 - 64 - 44 = 0

36(x - 1)² + 4(y + 4)² - 144 = 0

36(x - 1)² + 4(y + 4)² = 144

Dividing by 144 on both sides

(x - 1)²/4 + (y + 4)²/36 = 1

Let X = x - 1 and Y = y + 4

X²/4 + Y²/36 = 1

The ellipse is symmeteric about y-axis.

a² = 36 and b² = 4

a = 6 and b = 2

Center :

Eccentricity :

e = √1 - (b²/a²)

e = √1 - (4/36)

= √(36 - 4)/36

= √32/36

= 4√2/6

= 2√2/3

ae = 6(2√2/3)

= 4√2

Foci :

F (0, ae) F'(0, -ae)

Referred to X and Y :

F (0, 4√2) F'(0, -4√2)

Referred to x and y :

Vertices :

A(0, b) A'(o, -b)

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

Referred to X and Y :

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