# Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse x^{2}/100 + y^{2}/400 = 1

**Solution:**

The given equation is x^{2}/100 + y^{2}/400 or x^{2}/(10)^{2 }+ y^{2}/(20)^{2} = 1

Here, the denominator of y^{2}/400 is greater than the denominator of x^{2}/100

Therefore, the major axis is along the y-axis, while the minor axis is along the x-axis.

On comparing the given equation with

x^{2}/b^{2} + y^{2}/a^{2} = 1 we obtain b = 10 and a = 20

Hence,

c = √400² - 100²

= √300

= 10√3

Therefore,

The coordinates of the foci are (0, ± 10√3)

The coordinates of the vertices are (0 ± 20)

Length of major axis = 2a = 40

Length of minor axis = 2b = 20

Eccentricity, e = c/a = 10√3/20 = √3/2

Length of latus rectum = 2b^{2}/a = (2 × 100)/20 = 10

NCERT Solutions Class 11 Maths Chapter 11 Exercise 11.3 Question 6

## Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse x^{2}/100 + y^{2}/400 = 1

**Summary:**

The coordinates of the foci and vertices are (0, ± 10√3), (0 ± 20) respectively. The length of the major axis, minor axis, and latus rectum are 40, 20, 10 respectively