If e and e- are the eccentricities of the hyperbola x2-a2-y2-b2-1 and its conjugate hyperbola the value of 1-e2-1-e2 is

The correct option is B 1b^2 = a^2 (e^2 1) ⇒ e^2 1 = b^2/a^2 a^2 = b^2 (e^`2 1) ⇒ e^`2 1 = a^2/b^2 = 1/e^2 1⇒ (e^2 1)(e^`2 1) = 1 ⇒ e^2 e^`2 = e^ ...

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Byju's Answer

Standard XII

Mathematics

Definition of Hyperbola

If e and e' a...

Question

If e and e’ are the eccentricities of the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ and its conjugate hyperbola the value of $\frac{1}{e^{2}} + \frac{1}{e^{‘ 2}}$ is

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Solution

The correct option is C 1
$b^{2} = a^{2} (e^{2} - 1) \Rightarrow e^{2} - 1 = \frac{b^{2}}{a^{2}} a^{2} = b^{2} (e^{‘ 2} - 1) \Rightarrow e^{‘ 2} - 1 = \frac{a^{2}}{b^{2}} = \frac{1}{e^{2} - 1} \Rightarrow (e^{2} - 1) (e^{‘ 2} - 1) = 1 \Rightarrow e^{2} e^{‘ 2} = e^{2} + e^{‘ 2} \Rightarrow \frac{1}{e^{2}} + \frac{1}{e^{‘ 2}} = 1$

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Similar questions

Q. If the eccentricities of the hyperbola

\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1

and

\frac{y^{2}}{b^{2}} - \frac{x^{2}}{a^{2}} = 1

e

and

e_{1}

, then

\frac{1}{e^{2}} + \frac{1}{e_{1}^{2}} =

Q. If

e, e_{1}

are the eccentricities of hyperbola and its conjugate hyperbola prove that

\frac{1}{e^{2}} + \frac{1}{e_{1}^{2}} = 1

Q. If

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\frac{1}{e^{2}} + \frac{1}{e^{2}}

is equal to.

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\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

and hyperbola

\frac{x^{2}}{A^{2}} - \frac{y^{2}}{B^{2}} = 1

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\frac{1}{e_{1}^{2}} + \frac{1}{e_{2}^{2}}

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If e and e’ are the eccentricities of the hyperbola x2a2−y2b2=1 and its conjugate hyperbola the value of 1e2+1e‘2 is

The correct option is C 1b2=a2(e2−1)⇒e2−1=b2a2a2=b2(e‘2−1)⇒e‘2−1=a2b2=1e2−1⇒(e2−1)(e‘2−1)=1⇒e2 e‘2=e2+e‘2⇒1e2+1e‘2=1

If e and e’ are the eccentricities of the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ and its conjugate hyperbola the value of $\frac{1}{e^{2}} + \frac{1}{e^{‘ 2}}$ is