NIMCET 2012 Question Paper with solutions

Question : 36 of 120

Marks: +1, -0

If e and e' be the eccentricities of a hyperbolaand its conjugate, then

1/e^2+1/e^{'e}

is equal to

Solution:

We know that, the eccentricity of hyperbola is

b^2 = a^2(e^2-1)

⇒

b^2/a^2 = e^2

- 1
⇒

e^2 = {a^2+b^2}/a^2

⇒

1/e^2 = a^2/{a^2+b^2}

... (i)
and the eccentricity of its conjugate

a^2 = b^2(e'^2-1)

⇒

a^2/b^2 = e'^2

- 1
⇒

e'^2 = {a^2+b^2}b^2

⇒

1/{e'^2} = b^2/{a^2+b^2}

... (ii)
On adding Eqs. (i) and (ii), we get

1/e^2+1/{e'^2} = a^2/{a^2+b^2}+b^2/{a^2+b^2} = {a^2+b^2}/{a^2+b^2}

⇒

1/e^2+1/{e'^2} = 1

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