Question

If $e\,\,and\,\,{e_1}$ are the eccentricities of the hyperbola $xy = {c^2}$ and ${x^2} - {y^2} = {c^2}$ then ${e^2} + e_1^2 = ?$
A. 1
B. 4
C. 6
D. 8

Answer 1

Hint: First of all we will find the value of eccentricity of the given hyperbola. If we consider a hyperbola of the form, $\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1$ then, the eccentricity $= \dfrac{{\sqrt {{a^2} + {b^2}} }}{a}$.

Complete step-by-step solution -
Also, we know that if ${\rm{a}} = {\rm{b}}$ then it is known as a rectangular hyperbola. So, by using this property and formula we will get the required value.
We have been given that $e\,\,and\,\,{e_1}$are the eccentricities of the hyperbola $xy = {c^2}$ and ${x^2} - {y^2} = {c^2}$ respectively then we have to find ${e^2} + e_1^2$.
We know that eccentricity of a hyperbola $\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1$ is given by,
$e = \dfrac{{\sqrt {{a^2} + {b^2}} }}{a}$
Now, we know that a rectangular hyperbola is a hyperbola in which${\rm{a}} = {\rm{b}}$.
For $xy = {c^2}\,,\,\,a = b = c$
$\begin{array}{l} \Rightarrow e = \dfrac{{\sqrt {{c^2} + {c^2}} }}{c}\\\ \Rightarrow e = \dfrac{{\sqrt {2{c^2}} }}{c}\end{array}$
Since we know that square root of ${c^2}$ is c, we get:
$\Rightarrow e = \dfrac{{\sqrt 2 c}}{c}$
Cancelling c from numerator and denominator, we have the value of e as:
$\Rightarrow e = \sqrt 2$
Again, for ${x^2} - {y^2} = {c^2}\,\,,\,\,a = b = c$
$\begin{array}{l} \Rightarrow e = \dfrac{{\sqrt {{c^2} + {c^2}} }}{c}\\\ \Rightarrow e = \dfrac{{\sqrt {2{c^2}} }}{c}\end{array}$
Since we can also express the root of ${c^2}$as c. So, using this, we get:
$\Rightarrow e = \dfrac{{\sqrt 2 c}}{c}$
Now, c is common in numerator and denominator, so cancelling them we get:
$\Rightarrow e = \sqrt 2$
We have $e = \sqrt 2 \,\,and\,\,{e_1} = \sqrt 2$
So the value of ${e^2} + e_1^2 = {\left( {\sqrt 2 } \right)^2} + {\left( {\sqrt 2 } \right)^2}$
$\begin{array}{l} \Rightarrow 2 + 2\\\ \Rightarrow 4\end{array}$
Therefore, the correct option is B.

Note: If you remember the property that eccentricity of any rectangular hyperbola is always constant and it is equal to $\sqrt 2$ , it will save your time in this type of question. And a hyperbola in the form of $\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1$ where ${\rm{a}} = {\rm{b}}$ is known as rectangular hyperbola.