# Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola 49y^{2} - 16x^{2} = 784

**Solution:**

The given equation is 49y^{2} - 16x^{2} = 784

It can be written as

49y^{2} - 16x^{2} = 784

y^{2}/16 - x^{2}/49 = 1

y^{2}/4^{2} - x^{2}/7^{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 = 4 and b = 7.

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

Hence,

⇒ c^{2} = (4)^{2} + (7)^{2}

⇒ c^{2} = 16 + 49

⇒ c^{2} = 65

⇒ c = √65

Therefore,

The coordinates of the foci are (0, ± √65)

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

Eccentricity, e = c/a = √65/4

Length of latus rectum = 2b^{2}/a = (2 × 49)/4 = 49/2

NCERT Solutions Class 11 Maths Chapter 11 Exercise 11.4 Question 6

## Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola 49y^{2} - 16x^{2} = 784

**Summary:**

The coordinates of the foci and vertices of the hyperbola 49y^{2} - 16x^{2} = 784 are (0, ± √65), (0, ± 4) respectively. The length of the latus rectum is 49/2