# Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola y^{2}/9 - x^{2}/27 = 1

**Solution:**

The given equation is

y^{2}/9 - x^{2}/27 = 1

On comparing this equation with the standard equation of hyperbola

i.e., y^{2}/a^{2} - x^{2}/b^{2} = 1, we obtain

a = 3 and b = √27.

We know that, c^{2} = a^{2} + b^{2}

Hence,

⇒ c^{2} = 3^{2} + (√27)^{2}

⇒ c^{2} = 9 + 27

⇒ c^{2} = 36

⇒ c = 6

Therefore,

The coordinates of the foci are (0, ± 6)

The coordinates of the vertices are (0, ± 3)

Eccentricity, e = c/a = 6/3 = 2

Length of latus rectum = 2b^{2}/a = (2 × 27)/3 = 18

NCERT Solutions Class 11 Maths Chapter 11 Exercise 11.4 Question 2

## Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola y^{2}/9 - x^{2}/27 = 1

**Summary:**

The coordinates of the foci and vertices of the hyperbola y^{2}/9 - x^{2}/27 = 1 are (0, ± 6), (0, ± 3) respectively. The length of the latus rectum is 18

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