Name: Tangent to a Hyperbola
Uploaded: 2019-07-19
Duration: 2 min 11 s
Description: The tangent to a plane curve at a given point is the straight line that "just touches" the curve at that point. Learn about the tangent to a hyperbola by wat...

Question

If

e

and

e^{'}

are the eccentricities of a hyperbola and its conjugate hyperbola respectively, then prove that

\frac{1}{e^{2}} + \frac{1}{{(e^{'})}^{2}} = 1

.

Accepted Answer

Let the equation of hyperbola be $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ .

Therefore, the eccentricity of the hyperbola is $e = \sqrt{1 + {(\frac{b}{a})}^{2}}$ {From $b^{2} = a^{2} (e^{2} - 1)$ }

$\Rightarrow e^{2} = 1 + {(\frac{b}{a})}^{2} \to (1)$

And also, the conjugate hyperbola of $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ is $- \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ .

Therefore, the eccentricity of the conjugate hyperbola is $e^{'} = \sqrt{1 + \frac{a^{2}}{b^{2}}}$ {From $a^{2} = b^{2} ({(e^{'})}^{2} - 1)$ }

$\Rightarrow {(e^{'})}^{2} = 1 + \frac{a^{2}}{b^{2}} \to (2)$

Hence, from $(1)$ and $(2)$ , we can write $\frac{1}{e^{2}} + \frac{1}{{(e^{'})}^{2}} = \frac{1}{1 + \frac{b^{2}}{a^{2}}} + \frac{1}{1 + \frac{a^{2}}{b^{2}}}$

$∴ \frac{1}{e^{2}} + \frac{1}{{(e^{'})}^{2}} = \frac{a^{2}}{a^{2} + b^{2}} + \frac{b^{2}}{a^{2} + b^{2}}$

$∴ \frac{1}{e^{2}} + \frac{1}{{(e^{'})}^{2}} = \frac{a^{2} + b^{2}}{a^{2} + b^{2}}$

$∴ \frac{1}{e^{2}} + \frac{1}{{(e^{'})}^{2}} = 1$

Hence, proved.

If $e$ and $e^{'}$ are the eccentricities of a hyperbola and its conjugate hyperbola respectively, then prove that $\frac{1}{e^{2}} + \frac{1}{{(e^{'})}^{2}} = 1$ .

Important Questions on Conic Sections

Learn and Prepare for any exam you want !

If e and e' are the eccentricities of a hyperbola and its conjugate hyperbola respectively, then prove that 1e2+1e'2=1.

Important Questions on Conic Sections

Learn and Prepare for any exam you want !

If $e$ and $e^{'}$ are the eccentricities of a hyperbola and its conjugate hyperbola respectively, then prove that $\frac{1}{e^{2}} + \frac{1}{{(e^{'})}^{2}} = 1$ .