# Find the equation of the hyperbola satisfying the given conditions: Vertices (± 7, 0), e = 4/3.

**Solution:**

Vertices (± 7, 0),

e = 4/3.

Here, the foci are on the x-axis

Therefore,

the equation of the hyperbola is of the form x^{2}/a^{2} - y^{2}/b^{2} = 1

Since the vertices are (± 7, 0), a = 7

It is given that e = 4/3

Hence,

⇒ e = c/a = 4/3

⇒ c/7 = 4/3

⇒ c = 28/3

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

Therefore,

⇒ 7^{2} + b^{2} = (28/3)^{2}

⇒ b^{2} = 784/9 - 49

⇒ b^{2} = (784 - 441)/9

⇒ b^{2} = 343/9

Thus, the equation of the hyperbola is x^{2}/49 - 9y^{2}/343 = 1

NCERT Solutions Class 11 Maths Chapter 11 Exercise 11.4 Question 14

## Find the equation of the hyperbola satisfying the given conditions: Vertices (± 7, 0), e = 4/3

**Summary:**

The equation of the hyperbola is x^{2}/49 - 9y^{2}/343 = 1 while the vertices are (± 7, 0) and the value of e is 4/3