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

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