Question

Find the equation of the hyperbola of given length of transverse axis ' 6 ' whose vertex bisects the distance between the center and the focus.

Answer 1

Step-by-step explanation:

Let the equation of the Hyperbola be :

• x²/a² - y²/b² = 1

Let c is the centre, A is one vertex, S is the corresponding focus of the Hyperbola given length of transverse axis = 2a = 6

Given vertex bisects the distance between centre and focus.

• 2CA = CS
• 2a = ae
• e = 2

Since b² = a²(e² - 1)

• (3²)(2² - 1)
• 9(4 - 1)
• 9(3)
• 27

⇒ b² = 27

• b = √27
• b = 3√3

∴ Equation of hyperbola :

• x²/a² - y²/b² = 1

That is :

• x²/9 - y²/27 = 1

Answer 2

Standard equation of hyperbola is

a

2

x

2

−

b

2

y

2

=1

x-axis is the transverse axis

So, centre c=(0,0)

Vertex A=(a,0)

Focus S=(ae,0)

Vertex bisects the distance between centre and focus

i.e, vertex is the mid point of line joining vertex and focus

Applying mid point formula

a=

2

0+ae

2a=ae

e=2

We know that,

e

2

=1+

a

2

b

2

3=

a

2

b

2

Hence the equation of parabola in terms of a is

a

2

x

2

−

3a

2

y

2

=1