Question

Find the equation of hyperbola whose vertices are (±7,0) and the eccentricity is 4/3

Answer 1

Answer:

x2/49 – (9y2)/343 = 1

Step-by-step explanation:

Vertices (±7, 0), e = 4/3

Here, the vertices 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 , we know e = c/a

c/a = 4/3

c/7 = 4/3

c = 28/3

We know that a2 + b2 = c2.

72 + b2 = (28/3)2

b2 = 784/9 – 49

b2 = (784 – 441)/9 = 343/9

Thus the equation of the hyperbola

x2/49 – (9y2)/343 = 1

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Answer 2

$\Large{\textbf{\underline{\underline{According\:to\:the:Question}}}}$

Vertices of hyperbola are form (±a , 0)

Hence,

It is a horizontal ellipse

Now,

Assume

${\boxed{\sf\:{Equation=\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1}}}$

Vertices = (±a , 0)

But,

Vertices = (±7 , 0) [Given]

Therefore,

a = 7

a² = 49

$\tt{\rightarrow e=\dfrac{c}{a}}$

c = ae

$\tt{\rightarrow 7\times\dfrac{4}{3}}$

$\tt{\rightarrow\dfrac{28}{3}}$

b² = (c² - a²)

$\tt{\rightarrow(\dfrac{28}{3})^2-49}$

$\tt{\rightarrow\dfrac{343}{9}}$

Then,

$\tt{\rightarrow a^2=49\;and\;b^2=\dfrac{343}{9}}$

Hence,

$\Large{\boxed{\sf\:{Equation=\dfrac{x^2}{49}-\dfrac{y^2}{343/9}=1}}}$

$\Large{\boxed{\sf\:{Equation=\dfrac{x^2}{49}-\dfrac{9y^2}{343}=1}}}$

7x² - 9y² = 343