# 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 36x^{2} + 4y^{2} = 144

**Solution:**

The given equation is

36x^{2} + 4y^{2} = 144

It can be written as,

36x^{2} + 4y^{2} = 144

⇒ x^{2}/4 + y^{2}/36 = 1

⇒ x^{2}/(2)^{2} + y^{2}/(6)^{2} = 1

Here, the denominator of y^{2 }/ (6)^{2} is greater than the denominator of x^{2 }/ (2)^{2}

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 = 2 and a = 6

Hence,

c = √36 - 4

= √32

= 4√2

Therefore,

The coordinates of the foci are (0, ± 4√2)

The coordinates of the vertices are (0 ± 6)

Length of major axis = 2a = 12

Length of minor axis = 2b = 4

Eccentricity, e = c/a = 4√2/6 = 2√2/3

Length of latus rectum = 2b^{2}/a = (2 × 4)/6 = 4/3

NCERT Solutions Class 11 Maths Chapter 11 Exercise 11.3 Question 7

## 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 36x^{2} + 4y^{2} = 144

**Summary:**

The coordinates of the foci and vertices are (0, ± 4√2), (0 ± 6) respectively. The length of the major axis, minor axis, and latus rectum are 12, 4, 4/3 respectively.

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