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 x²/a² - y²/b² = 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, c² = a² + b²

Therefore,

⇒ 7² + b² = (28/3)²

⇒ b² = 784/9 - 49

⇒ b² = (784 - 441)/9

⇒ b² = 343/9

Thus, the equation of the hyperbola is x²/49 - 9y²/343 = 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 x²/49 - 9y²/343 = 1 while the vertices are (± 7, 0) and the value of e is 4/3