FIND THE EQUATION OF THE ELLIPSE WITH THE GIVEN INFORMATION

About "Find the Equation of the Ellipse with the Given Information"

Find the Equation of the Ellipse with the Given Information :

Here we are going to see some practice questions on find the equation of the ellipse with the given information.

Find the Equation of the Ellipse with the Given Information - Practice questions

Question 1 :

Find the equation of the ellipse in each of the cases given below:

(i) foci (± 3 0), e  =  1/2

Solution :

F₁ (3, 0) and F₂ (-3, 0) and e = 1/2

From the given information, we know that the given ellipse is symmetric about x axis.

Midpoint of foci  =  Center of the ellipse

Center  =  (3 + (-3))/2, (0 + 0)/2  =  C (0, 0)

Distance between foci  =  √(x₂ - x₁)² + (y₂ - y₁)²

  =  √(3 + 3)² + (0 - 0)²

  =  √6² + (0 - 0)²

2ae  =  6

ae  =  3

a(1/2)  =  3

a  =  6

b²  =  a² (1 - e²)

b²  =  6² (1 - (1/2)²)

b²  =  36 (3/4)

b²  =  27

(x²/a²) + (y²/b²)  =  1

(x²/36) + (y²/27)  =  1

(ii) foci (0, ± 4) and end points of major axis are (0, ± 5).

Solution :

F₁ (0, 4) and F₂ (0, -4)

From the given foci, we know that the ellipse is symmetric about y-axis.

Distance between foci  =  √(x₂ - x₁)² + (y₂ - y₁)²

2ae  =  √(0 - 0)² + (4 + 4)²

2ae  =  8

ae  =  4

Distance between end points of major axis

  =  √(0 - 0)² + (5 + 5)²

2a  =  10

a  =  5

5e  =  4

e  =  4/5

b²  =  a²(1 - e²)

b²  =  5² (1 - (4/5)²)

b²  =  5² (9/25)

b²  =  9

Hence the required equation of ellipse is 

(x²/25) + (y²/9)  =  1

(iii)  length of latus rectum 8, eccentricity = 3/5 and major axis on x -axis.

Solution :

Length of latus rectum  =  8

2b²/a  =  8

b²  =  4a

e  =  3/5

Since the major axis is on x-axis, the ellipse is symmetric about x-axis.

b²  =  a² (1 - e²)

4a  =  a² (1 - (3/5)²)

4  =  a (16/25)

a  =  25/4

b²  =  4(25/4)

b²  =  25

(x²/(16/625)) + (y²/25)  =  1

(625x²/16) + (y²/25)  =  1

(iv) length of latus rectum 4 , distance between foci 4√2 and major axis as y - axis.

Solution :

length of latus rectum = 4

2b²/a  =  4

b²  =  2a -------(1) 

Distance between foci  =  4√2 

2ae  =  4√2 

ae  =  2√2 

b²  =  a² (1 - e²)

b²  =  a² - (ae)²

b²  =  a² - (2√2)²

b²  =  a² - 8 -------(2)

2a  =  a² - 8

a² - 2a  - 8  =  0

(a - 4) (a + 2)  =  0

a  =  4 and a = -2

If a = 4, then b²  =  2(4)  =  8

If a = -2, then b²  =  2(-2)  =  -4 (Not admissible)

(x²/8) + (y²/16)  =  1

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