# Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola x^{2}/16 - y^{2}/9 = 1

**Solution:**

The given equation is x^{2}/16 - y^{2}/9 = 1

On comparing this equation with the standard equation of hyperbola

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

a = 4 and b = 3.

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

Hence,

⇒ c^{2} = 4^{2} + 3^{2}

⇒ c^{2} = 25

⇒ c = 5

Therefore,

The coordinates of the foci are (± 5, 0)

The coordinates of the vertices are (± 4, 0)

Eccentricity, e = c/a = 5/4

Length of latus rectum = 2b^{2}/a = (2 × 9)/4 = 9/2

NCERT Solutions Class 11 Maths Chapter 11 Exercise 11.4 Question 1

## Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola x^{2}/16 - y^{2}/9 = 1

**Summary:**

The coordinates of the foci and vertices of the hyperbola x^{2}/16 - y^{2}/9 = 1 are (± √5, 0), (± 4, 0) respectively. The length of the latus rectum is 9/2