If the eccentricities of the hyperbolas x2-a2-y2-b2-1 and y2-b2-x2-a2-1 be e and e1- then 1-e2-1-e12-A- 1B- 2C- 3D- None of these

The correct options are A 1 B 2 e=√(1+b^2 a^2) ⇒ e^2=a^2+b^2/a^2 e_1=√(1+a^2 b^2) ⇒ e_1^2=b^2+a^2/b^2 ⇒1 e_1^2+1 e^2=1 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XI

Mathematics

Sum of binomial coefficients with alternate signs

If the eccent...

Question

If the eccentricities of the hyperbolas $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 a n d \frac{y^{2}}{b^{2}} - \frac{x^{2}}{a^{2}} = 1$ be e and $e_{1}$ , then $\frac{1}{e^{2}} + \frac{1}{e_{1}^{2}} =$

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

None of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct options are
A
1

B
2

$e = \sqrt{1 + \frac{b^{2}}{a^{2}}} \Rightarrow e^{2} = \frac{a^{2} + b^{2}}{a^{2}}$
$e_{1} = \sqrt{1 + \frac{a^{2}}{b^{2}}} \Rightarrow e_{1}^{2} = \frac{b^{2} + a^{2}}{b^{2}} \Rightarrow \frac{1}{e_{1}^{2}} + \frac{1}{e^{2}} = 1$

Suggest Corrections

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}} =

If $e_{1} a n d e_{2}$ are the eccentricity of hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ and $\frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1$ , then point $\frac{1}{e_{1}}, \frac{1}{e_{2}}$ lies on the circle.

Q. If eccentricities of two hyperbolas

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

and

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

are

e_{1}

and

e_{2}

respectively then prove that

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

Q. The ellipse

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

and hyperbola

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

are having a same foci and length of minor axis of ellipse is same as the conjugate axis of the hyperbola. If

e_{1}

e_{2}

are the eccentricities of ellipse and hyperbola respectively, then the value of

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

Q. If the eccentricity of the hyperbola

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

e

then the eccentricity of the hyperbola

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

is :

Join BYJU'S Learning Program

If the eccentricities of the hyperbolas x2a2−y2b2=1 and y2b2−x2a2=1 be e and e1, then 1e2+1e21=

The correct options are A 1 B 2 e=√1+b2a2 ⇒e2=a2+b2a2 e1=√1+a2b2 ⇒e21=b2+a2b2 ⇒1e21+1e2=1

If the eccentricities of the hyperbolas $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 a n d \frac{y^{2}}{b^{2}} - \frac{x^{2}}{a^{2}} = 1$ be e and $e_{1}$ , then $\frac{1}{e^{2}} + \frac{1}{e_{1}^{2}} =$

The correct options are
A
1

B
2

$e = \sqrt{1 + \frac{b^{2}}{a^{2}}} \Rightarrow e^{2} = \frac{a^{2} + b^{2}}{a^{2}}$
$e_{1} = \sqrt{1 + \frac{a^{2}}{b^{2}}} \Rightarrow e_{1}^{2} = \frac{b^{2} + a^{2}}{b^{2}} \Rightarrow \frac{1}{e_{1}^{2}} + \frac{1}{e^{2}} = 1$