0

Guest

Courses New

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
3 years 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

Think You Can Provide A Better Answer ?

Other Related Questions on Analytical Geometry

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More