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Determine the foci coordinates, the vertices, the length of the major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse (x 2 /49) + (y 2 /36) = 1

Determine the foci coordinates, the vertices, the length of the major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse (x2/49) + (y2/36) = 1

Grade:12

1 Answers

Harshit Singh
askIITians Faculty 5963 Points
9 months ago
Welcome to AskIITians

The given equation is (x^2/49) + (y^2/36) = 1

It can be written as (x^2/7^2) + (y^2/6^2) = 1

It is noticed that the denominator of x^2/49 is greater than the denominator of the y^2/36

On comparing the equation with (x^2/a^2) + (y^2/b^2) = 1, we will get

a= 7 and b = 6

Therefore, c = √(a^2– b^2)

Now, substitute the value of a and b

⇒ √(a^2– b^2) = √(72– 62) = √(49-36)

⇒ √13

Hence, the foci coordinates are ( ± √13, 0)

Eccentricity, e = c/a = √13/ 7

Length of the major axis = 2a = 2(7) = 14

Length of the minor axis = 2b = 2(6) =12

The coordinates of the vertices are ( ± 7, 0)

Latus rectum Length= 2b^2/a = 2(6)^2/7 = 2(36)/7 = 72/7

Thanks

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