If e 1 and e 2 are the eccentricity of hyperbola x 2- a 2- y 2- b 2-1 and y 2- a 2- x 2- b 2-1- then point 1- e 1- 1- e 2 lies on the circle-A- x2 y2-1B- x2 y2-2C- x2 y2-3D- x2 y2-4

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Byju's Answer

Standard XII

Mathematics

Equation of Tangent at a Point (x,y) in Terms of f'(x)

If e 1 and e ...

Question

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.

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Solution

The correct option is A

We already discussed the property

$e_{1}^{- 2} + e_{2}^{- 2} = 1$ in a given hyperbola and its conjugate hyperbola.

This is a different way of asking the same property.

Since (x,y)= $(e_{1}^{- 2} + e_{2}^{- 2} = 1)$

$x^{2} + y^{2} = 1$ will be circle on which the point would always fall due to the property.

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Similar questions

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

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. 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. 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. Let the curves

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 4

and

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

have same foci. If the eccentricity of the given curves are

e_{1}

and

e_{2}

respectively, then

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If e1 and e2 are the eccentricity of hyperbola x2a2−y2b2=1 and y2a2−x2b2=1 , then point 1e1,1e2 lies on the circle.

The correct option is A We already discussed the property e−21+e−22=1 in a given hyperbola and its conjugate hyperbola. This is a different way of asking the same property. Since (x,y)=(e−21+e−22=1) x2+y2=1 will be circle on which the point would always fall due to the property.

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.