FIND THE EQUATION OF AN ELLIPSE FROM FOCI AND ECCENTRICITY

Example 1 :

Find the equation of the ellipse whose focus is (-1, 1), eccentricity is 1/2 and whose directrix is x-y+3  =  0.

Solution :

Let P(x, y) be the fixed point on ellipse.

F(-1, 1) and M is directrix.

By definition :

FP/PM  =  e

FP  =  e(PM)

√(x+1)² + (y-1)²   =  (1/2) [(x-y+3)/√1²+(-1)²]

Taking squares on both sides, we get

(x+1)² + (y-1)²   =  (1/4) [(x-y+3)²/2]

x²+2x+1+y²-2y+1  =  (1/8) (x²+y²+9-2xy-6y+6x)

8(x²+2x+1+y²-2y+1)  =  x²+y²+9-2xy-6y+6x

8x²-x²+2xy+8y²-y²+16x-6x-16y+6y+8-9  =  0

7x²+2xy+7y²+10x-10y-1 =  0

Example 2 :

Find the equation of the ellipse whose foci are (2, 0) and (-2, 0) and eccentricity is 1/2.

Solution :

Midpoint of foci  =  Center

Midpoint  =  (x₁+x₂)/2, (y₁+y₂)/2

  =  (2-2)/2, (0+0)/2

Center  =  (0, 0)

Distance between center and foci  =  ae

=  √(0-2)² + (0-0)² 

ae  =  2

here e  =  1/2

a  =  4, a²  =  16

b²  =  a²(1-e²)

b²  =  16(1-(1/4))

b²  =  12

The ellipse is symmetric about x axis and center is (0, 0).

x²/a² + y²/b²  =  1

x²/1 + y²/12  =  1

Example 3 :

Find the equation of the ellipse whose foci are (2, -1) and (0, -1) and eccentricity is 1/2.

Solution :

Midpoint of foci  =  Center

Midpoint  =  (x₁+x₂)/2, (y₁+y₂)/2

  =  (2+0)/2, (-1-1)/2

  =  2/2, -2/2

=  (1, -1)

Center  =  (1, -1)

Distance between center and foci  =  ae

F₁ (2, -1) and C (1, -1)

=  √(2-1)² + (-1+1)² 

ae  =  1

here e  =  1/2

a  =  2

a²  =  4

b²  =  a²(1-e²)

b²  =  4(1-(1/4))

b²  =  3

The ellipse is symmetric about x-axis.

(x-1)²/4 + (y+1)²/3  =  1

Example 4 :

The center is (3, − 4), one of the foci is (3+√3, − 4) and

e = √3/2.

Solution :

Distance between center and foci  =  ae

ae  =  √(3-3-√3)² + (-4+4)² 

ae  =  √(3+0)

ae  =  √3

e  =  √3/2

a  =  2

a²  =  4

b²  =  a²(1-e²)

b²  =  4(1-(3/4))

b²  =  1

The ellipse is symmetric about x-axis.

(x-3)²/4 + (y+4)²/1  =  1

Example 5 :

the center at the origin, the major axis is along x-axis, e = 2/3 and passes through the point (2, -5/3).

Solution :

From the given information, the ellipse is symmetric about x-axis and center (0, 0)

x²/a²+y²/b²  =  1

The ellipse passes through the point (2, -5/3)

4//a²+ (25/9)/b²  =  1  --(1)

b²  =  a²(1-e²)

Applying the value of e, we get

b²  =  a²(1-(4/9))

b²  =  5a²/9

By applying the value of b² in (1), we get

4/a²+ (25/9)/(5a²/9)  =  1 

9/a²  =  1

a²  =  9

Applying the value of a², we get

b²  =  5

x²/9+y²/5  =  1

Find the equation of ellipse in standard form with the following information.

Example 6 :

Major axis horizontal with length 8, length of minor axis 4 and center (0, 0).

Solution :

Major axis is horizontal axis, then the ellipse is symmetric about x-axis.

Length of major axis = 8

2a = 8, a = 4

length of minor axis = 4

2b = 4, b = 2

x²/4² + y²/2²  =  1

x²/16 + y²/4  =  1

Example 7 :

Find the standard form of an equation of ellipse when foci are (±2, 0) and length of major axis is 12.

Solution :

Distance between foci = 2ae

2ae = 2 ----(1)

Length of major axis = 12

2a = 12

a = 6

Applying the value of a in (1), we get

2(6) e = 2

e = 2/12

e = 1/6

b² = a² (1 - e²)

b² = 6² (1 - (1/6)²)

= 36 (35/36)

b² = 35

From the given information, the ellipse is symmetric about x-axis and its center is (0, 0).

x² / a² + y² / b² = 1

x² / 36 + y² / 35 = 1

Example 8 :

Find the standard form of an equation of ellipse when center is (2, -1), vertex (2, 1/2) and length of minor axis is 2.

Solution :

By observing the given points, the ellipse is symmetric about y-axis. 

Distance between center and vertex 

b = √(y₂ - y₁)² + (x₂ - x₁)²

= √(1/2 + 1)² + (2 - 2)²

= √(3/2)² + 0²

= √(9/4)

b = 3/2

Length of minor axis a = 2

Equation of ellipse which is symmetric about y-axis with center (h, k).

(x - h)² / a² + (y - k)² / b² = 1

(x - 2)² / 2² + (y + 1)² / (3/2)² = 1

(x - 2)² / 4 + (y + 1)² / (9/4) = 1

Example 9 :

Find the standard form of an equation of ellipse when center is (3, 2), a = 3c and foci are (1, 2) and (5, 2).

Solution :

Distance between foci = 2ae

2ae = √(y₂ - y₁)² + (x₂ - x₁)²

= √(2 - 2)² + (5 - 1)²

= √4²

2c = 4

c = 2

c² = a² - b²

a = 3c ==> a = 3(2) = 6

2² = 6² - b²

b² = 36 - 4

b² = 32

(x - 3)² / 36 + (y - 3)² / 32 = 1

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