# Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola 16x^{2} - 9y^{2} = 576

**Solution:**

The given equation is 16x^{2} - 9y^{2} = 576

It can be written as

16x^{2} - 9y^{2} = 576

x^{2}/ 36 - y^{2}/ 64 = 1

x^{2}/6^{2} - y^{2}/ 8^{2} = 1 ....(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 = 6 and b = 8.

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

Hence,

⇒ c^{2} = 6^{2} + 8^{2}

⇒ c^{2} = 36 + 64

⇒ c^{2} = 100

⇒ c = 10

Therefore,

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

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

Eccentricity, e = c/a = 10/6 = 5/3

Length of latus rectum = 2b^{2}/a = (2 × 64)/6 = 64/3

NCERT Solutions Class 11 Maths Chapter 11 Exercise 11.4 Question 4

## Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola 16x^{2} - 9y^{2} = 576.

**Solution:**

The coordinates of the foci and vertices of the hyperbola 16x^{2} - 9y^{2} = 576 are (± 10, 0), (± 6, 0) respectively. The length of the latus rectum is 64/3