If ${{\text{e}}_{\text{1}}}$ and ${{\text{e}}_{\text{2}}}$ are respectively, the eccentricity of a hyperbola and its conjugate. Prove that $\dfrac{1}{{{\text{e}}_1^2}} + \dfrac{1}{{{\text{e}}_2^2}} = 1$
Answer
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Hint: From the given question, we have to prove that the reciprocal of two eccentricities equal to $1$. Now, we have to prove this by finding the eccentricities ${{\text{e}}_{\text{1}}}$ and ${{\text{e}}_{\text{2}}}$ from the two hyperbolas and substituting in the left hand side (LHS) to get the right hand side (RHS).
The locus of a point whose distance from a fixed point bears a constant ratio, greater than one to its distance from a fixed line is called a hyperbola.
Complete step-by-step solution:
Let ${{\text{e}}_{\text{1}}}$ be the eccentricity of hyperbola \[\dfrac{{{{\text{x}}^{\text{2}}}}}{{{{\text{a}}^{\text{2}}}}} - \dfrac{{{{\text{y}}^{\text{2}}}}}{{{{\text{b}}^{\text{2}}}}} = 1\] where \[{{\text{b}}^2} = {{\text{a}}^{\text{2}}}\left( {{\text{e}}_1^{\text{2}} - 1} \right)\] and ${{\text{e}}_{\text{2}}}$be the eccentricities of hyperbola \[\dfrac{{{{\text{y}}^{\text{2}}}}}{{{{\text{b}}^{\text{2}}}}} - \dfrac{{{{\text{x}}^{\text{2}}}}}{{{{\text{a}}^{\text{2}}}}} = 1\] where \[{{\text{a}}^2} = {{\text{b}}^{\text{2}}}\left( {{\text{e}}_2^{\text{2}} - 1} \right)\].
Then the eccentricity of \[\dfrac{{{{\text{x}}^{\text{2}}}}}{{{{\text{a}}^{\text{2}}}}} - \dfrac{{{{\text{y}}^{\text{2}}}}}{{{{\text{b}}^{\text{2}}}}} = 1\] where \[{{\text{b}}^2} = {{\text{a}}^{\text{2}}}\left( {{\text{e}}_1^{\text{2}} - 1} \right)\] is \[{\text{e}}_{\text{1}}^{\text{2}} = \dfrac{{{{\text{b}}^2}}}{{{{\text{a}}^{\text{2}}}}} + 1\] and the eccentricity of \[\dfrac{{{{\text{y}}^{\text{2}}}}}{{{{\text{b}}^{\text{2}}}}} - \dfrac{{{{\text{x}}^{\text{2}}}}}{{{{\text{a}}^{\text{2}}}}} = 1\] where \[{{\text{a}}^2} = {{\text{b}}^{\text{2}}}\left( {{\text{e}}_2^{\text{2}} - 1} \right)\] is \[{\text{e}}_2^{\text{2}} = \dfrac{{{{\text{a}}^{\text{2}}}}}{{{{\text{b}}^{\text{2}}}}} + 1\] .
Now, we are going to prove that the $\dfrac{1}{{{\text{e}}_1^2}} + \dfrac{1}{{{\text{e}}_2^2}}$is equals to $1$
So, we have to substitute the eccentricities ${\text{e}}_{\text{1}}^{\text{2}}$ and \[\;{\text{e}}_{\text{2}}^{\text{2}}\] in the $\dfrac{1}{{{\text{e}}_1^2}} + \dfrac{1}{{{\text{e}}_2^2}}$ to prove that the answer be equals to $1$.
$\dfrac{1}{{{\text{e}}_1^2}} + \dfrac{1}{{{\text{e}}_2^2}} = \dfrac{1}{{\;\dfrac{{{{\text{b}}^{\text{2}}}}}{{{{\text{a}}^{\text{2}}}}} + 1}} + \dfrac{1}{{\;\dfrac{{{{\text{a}}^{\text{2}}}}}{{{{\text{b}}^{\text{2}}}}} + 1}}$
Now take Least Common Multiple (LCM) on the denominator of right hand side (RHS). Then we get,
$\dfrac{1}{{{\text{e}}_1^2}} + \dfrac{1}{{{\text{e}}_2^2}} = \dfrac{1}{{\;\dfrac{{{{\text{b}}^{\text{2}}}}}{{{{\text{a}}^{\text{2}}}}} + \dfrac{{{{\text{a}}^{\text{2}}}}}{{{{\text{a}}^{\text{2}}}}}}} + \dfrac{1}{{\;\dfrac{{{{\text{a}}^{\text{2}}}}}{{{{\text{b}}^{\text{2}}}}} + \dfrac{{{{\text{b}}^{\text{2}}}}}{{{{\text{b}}^{\text{2}}}}}}}$
Simplifying we get,
$ \Rightarrow \dfrac{1}{{{\text{e}}_1^2}} + \dfrac{1}{{{\text{e}}_2^2}} = \dfrac{1}{{\;\dfrac{{{{\text{b}}^{\text{2}}} + {{\text{a}}^{\text{2}}}}}{{{{\text{a}}^{\text{2}}}}}}} + \dfrac{1}{{\;\dfrac{{{{\text{a}}^{\text{2}}} + {{\text{b}}^{\text{2}}}}}{{{{\text{b}}^{\text{2}}}}}}}$
Hence,
$ \Rightarrow \dfrac{1}{{{\text{e}}_1^2}} + \dfrac{1}{{{\text{e}}_2^2}} = \dfrac{{{{\text{a}}^{\text{2}}}}}{{\;{{\text{b}}^{\text{2}}} + {{\text{a}}^{\text{2}}}}} + \dfrac{{{{\text{b}}^{\text{2}}}}}{{\;{{\text{a}}^{\text{2}}} + {{\text{b}}^{\text{2}}}}}$
Here, we see that the denominators are same. So, we have to add up the numerators. Then we get,
$ \Rightarrow \dfrac{1}{{{\text{e}}_1^2}} + \dfrac{1}{{{\text{e}}_2^2}} = \dfrac{{{{\text{a}}^{\text{2}}} + {{\text{b}}^{\text{2}}}}}{{{{\text{a}}^{\text{2}}} + {{\text{b}}^{\text{2}}}}}$
$ \Rightarrow \dfrac{1}{{{\text{e}}_1^2}} + \dfrac{1}{{{\text{e}}_2^2}} = 1$ .
Therefore, we proved that the $\dfrac{1}{{{\text{e}}_1^2}} + \dfrac{1}{{{\text{e}}_2^2}}$is equals to $1$.
Thus, left hand side (LHS) = right hand side (RHS).
Hence, proved the required result.
Note: We have mind that, The standard equation of the hyperbola is \[\dfrac{{{{\text{x}}^{\text{2}}}}}{{{{\text{a}}^{\text{2}}}}} - \dfrac{{{{\text{y}}^{\text{2}}}}}{{{{\text{b}}^{\text{2}}}}} = 1\] where, \[{{\text{b}}^2} = {{\text{a}}^{\text{2}}}\left( {{\text{e}}_1^{\text{2}} - 1} \right)\] .
Also, let ${{\text{e}}_{\text{1}}}$ be the eccentricity of hyperbola \[\dfrac{{{{\text{x}}^{\text{2}}}}}{{{{\text{a}}^{\text{2}}}}} - \dfrac{{{{\text{y}}^{\text{2}}}}}{{{{\text{b}}^{\text{2}}}}} = 1\] where, \[{{\text{b}}^2} = {{\text{a}}^{\text{2}}}\left( {{\text{e}}_1^{\text{2}} - 1} \right)\] .
Now, we are going to get the eccentricity ${{\text{e}}_{\text{1}}}$ from the term \[{{\text{b}}^2} = {{\text{a}}^{\text{2}}}\left( {{\text{e}}_1^{\text{2}} - 1} \right)\]
\[ \Rightarrow \dfrac{{{{\text{b}}^2}}}{{{{\text{a}}^{\text{2}}}}} = \left( {{\text{e}}_{\text{1}}^{\text{2}} - 1} \right)\]
Rearranging the terms,
\[ \Rightarrow {\text{e}}_{\text{1}}^{\text{2}} = \dfrac{{{{\text{b}}^2}}}{{{{\text{a}}^{\text{2}}}}} + 1\]
Hence,
$ \Rightarrow {{\text{e}}_{\text{1}}} = \sqrt {1 + \dfrac{{{{\text{b}}^{\text{2}}}}}{{{{\text{a}}^{\text{2}}}}}} $
If the transverse axis along the y-axis and the conjugate axis is along x-axis, then the equation of the hyperbola \[\dfrac{{{{\text{y}}^{\text{2}}}}}{{{{\text{b}}^{\text{2}}}}} - \dfrac{{{{\text{x}}^{\text{2}}}}}{{{{\text{a}}^{\text{2}}}}} = 1\] where \[{{\text{a}}^2} = {{\text{b}}^{\text{2}}}\left( {{\text{e}}_2^{\text{2}} - 1} \right)\] .
Let ${{\text{e}}_{\text{2}}}$ be the eccentricities of hyperbola \[\dfrac{{{{\text{y}}^{\text{2}}}}}{{{{\text{b}}^{\text{2}}}}} - \dfrac{{{{\text{x}}^{\text{2}}}}}{{{{\text{a}}^{\text{2}}}}} = 1\] where \[{{\text{a}}^2} = {{\text{b}}^{\text{2}}}\left( {{\text{e}}_2^{\text{2}} - 1} \right)\] .
Now, we are going to get ${{\text{e}}_{\text{2}}}$ from the term \[{{\text{a}}^2} = {{\text{b}}^{\text{2}}}\left( {{\text{e}}_2^{\text{2}} - 1} \right)\] .
\[ \Rightarrow \dfrac{{{{\text{a}}^2}}}{{{{\text{b}}^{\text{2}}}}} = \left( {{\text{e}}_2^{\text{2}} - 1} \right)\]
Rearranging the terms we get,
\[ \Rightarrow {\text{e}}_2^{\text{2}} = \dfrac{{{{\text{a}}^{\text{2}}}}}{{{{\text{b}}^{\text{2}}}}} + 1\]
Hence,
$ \Rightarrow {{\text{e}}_2} = \sqrt {1 + \dfrac{{{{\text{a}}^{\text{2}}}}}{{{{\text{b}}^{\text{2}}}}}} $
Let \[\dfrac{{{{\text{x}}^{\text{2}}}}}{{{{\text{a}}^{\text{2}}}}} - \dfrac{{{{\text{y}}^{\text{2}}}}}{{{{\text{b}}^{\text{2}}}}} = 1\] and \[\dfrac{{{{\text{y}}^{\text{2}}}}}{{{{\text{b}}^{\text{2}}}}} - \dfrac{{{{\text{x}}^{\text{2}}}}}{{{{\text{a}}^{\text{2}}}}} = 1\] be the two hyperbola conjugate to each other.
The locus of a point whose distance from a fixed point bears a constant ratio, greater than one to its distance from a fixed line is called a hyperbola.
Complete step-by-step solution:
Let ${{\text{e}}_{\text{1}}}$ be the eccentricity of hyperbola \[\dfrac{{{{\text{x}}^{\text{2}}}}}{{{{\text{a}}^{\text{2}}}}} - \dfrac{{{{\text{y}}^{\text{2}}}}}{{{{\text{b}}^{\text{2}}}}} = 1\] where \[{{\text{b}}^2} = {{\text{a}}^{\text{2}}}\left( {{\text{e}}_1^{\text{2}} - 1} \right)\] and ${{\text{e}}_{\text{2}}}$be the eccentricities of hyperbola \[\dfrac{{{{\text{y}}^{\text{2}}}}}{{{{\text{b}}^{\text{2}}}}} - \dfrac{{{{\text{x}}^{\text{2}}}}}{{{{\text{a}}^{\text{2}}}}} = 1\] where \[{{\text{a}}^2} = {{\text{b}}^{\text{2}}}\left( {{\text{e}}_2^{\text{2}} - 1} \right)\].
Then the eccentricity of \[\dfrac{{{{\text{x}}^{\text{2}}}}}{{{{\text{a}}^{\text{2}}}}} - \dfrac{{{{\text{y}}^{\text{2}}}}}{{{{\text{b}}^{\text{2}}}}} = 1\] where \[{{\text{b}}^2} = {{\text{a}}^{\text{2}}}\left( {{\text{e}}_1^{\text{2}} - 1} \right)\] is \[{\text{e}}_{\text{1}}^{\text{2}} = \dfrac{{{{\text{b}}^2}}}{{{{\text{a}}^{\text{2}}}}} + 1\] and the eccentricity of \[\dfrac{{{{\text{y}}^{\text{2}}}}}{{{{\text{b}}^{\text{2}}}}} - \dfrac{{{{\text{x}}^{\text{2}}}}}{{{{\text{a}}^{\text{2}}}}} = 1\] where \[{{\text{a}}^2} = {{\text{b}}^{\text{2}}}\left( {{\text{e}}_2^{\text{2}} - 1} \right)\] is \[{\text{e}}_2^{\text{2}} = \dfrac{{{{\text{a}}^{\text{2}}}}}{{{{\text{b}}^{\text{2}}}}} + 1\] .
Now, we are going to prove that the $\dfrac{1}{{{\text{e}}_1^2}} + \dfrac{1}{{{\text{e}}_2^2}}$is equals to $1$
So, we have to substitute the eccentricities ${\text{e}}_{\text{1}}^{\text{2}}$ and \[\;{\text{e}}_{\text{2}}^{\text{2}}\] in the $\dfrac{1}{{{\text{e}}_1^2}} + \dfrac{1}{{{\text{e}}_2^2}}$ to prove that the answer be equals to $1$.
$\dfrac{1}{{{\text{e}}_1^2}} + \dfrac{1}{{{\text{e}}_2^2}} = \dfrac{1}{{\;\dfrac{{{{\text{b}}^{\text{2}}}}}{{{{\text{a}}^{\text{2}}}}} + 1}} + \dfrac{1}{{\;\dfrac{{{{\text{a}}^{\text{2}}}}}{{{{\text{b}}^{\text{2}}}}} + 1}}$
Now take Least Common Multiple (LCM) on the denominator of right hand side (RHS). Then we get,
$\dfrac{1}{{{\text{e}}_1^2}} + \dfrac{1}{{{\text{e}}_2^2}} = \dfrac{1}{{\;\dfrac{{{{\text{b}}^{\text{2}}}}}{{{{\text{a}}^{\text{2}}}}} + \dfrac{{{{\text{a}}^{\text{2}}}}}{{{{\text{a}}^{\text{2}}}}}}} + \dfrac{1}{{\;\dfrac{{{{\text{a}}^{\text{2}}}}}{{{{\text{b}}^{\text{2}}}}} + \dfrac{{{{\text{b}}^{\text{2}}}}}{{{{\text{b}}^{\text{2}}}}}}}$
Simplifying we get,
$ \Rightarrow \dfrac{1}{{{\text{e}}_1^2}} + \dfrac{1}{{{\text{e}}_2^2}} = \dfrac{1}{{\;\dfrac{{{{\text{b}}^{\text{2}}} + {{\text{a}}^{\text{2}}}}}{{{{\text{a}}^{\text{2}}}}}}} + \dfrac{1}{{\;\dfrac{{{{\text{a}}^{\text{2}}} + {{\text{b}}^{\text{2}}}}}{{{{\text{b}}^{\text{2}}}}}}}$
Hence,
$ \Rightarrow \dfrac{1}{{{\text{e}}_1^2}} + \dfrac{1}{{{\text{e}}_2^2}} = \dfrac{{{{\text{a}}^{\text{2}}}}}{{\;{{\text{b}}^{\text{2}}} + {{\text{a}}^{\text{2}}}}} + \dfrac{{{{\text{b}}^{\text{2}}}}}{{\;{{\text{a}}^{\text{2}}} + {{\text{b}}^{\text{2}}}}}$
Here, we see that the denominators are same. So, we have to add up the numerators. Then we get,
$ \Rightarrow \dfrac{1}{{{\text{e}}_1^2}} + \dfrac{1}{{{\text{e}}_2^2}} = \dfrac{{{{\text{a}}^{\text{2}}} + {{\text{b}}^{\text{2}}}}}{{{{\text{a}}^{\text{2}}} + {{\text{b}}^{\text{2}}}}}$
$ \Rightarrow \dfrac{1}{{{\text{e}}_1^2}} + \dfrac{1}{{{\text{e}}_2^2}} = 1$ .
Therefore, we proved that the $\dfrac{1}{{{\text{e}}_1^2}} + \dfrac{1}{{{\text{e}}_2^2}}$is equals to $1$.
Thus, left hand side (LHS) = right hand side (RHS).
Hence, proved the required result.
Note: We have mind that, The standard equation of the hyperbola is \[\dfrac{{{{\text{x}}^{\text{2}}}}}{{{{\text{a}}^{\text{2}}}}} - \dfrac{{{{\text{y}}^{\text{2}}}}}{{{{\text{b}}^{\text{2}}}}} = 1\] where, \[{{\text{b}}^2} = {{\text{a}}^{\text{2}}}\left( {{\text{e}}_1^{\text{2}} - 1} \right)\] .
Also, let ${{\text{e}}_{\text{1}}}$ be the eccentricity of hyperbola \[\dfrac{{{{\text{x}}^{\text{2}}}}}{{{{\text{a}}^{\text{2}}}}} - \dfrac{{{{\text{y}}^{\text{2}}}}}{{{{\text{b}}^{\text{2}}}}} = 1\] where, \[{{\text{b}}^2} = {{\text{a}}^{\text{2}}}\left( {{\text{e}}_1^{\text{2}} - 1} \right)\] .
Now, we are going to get the eccentricity ${{\text{e}}_{\text{1}}}$ from the term \[{{\text{b}}^2} = {{\text{a}}^{\text{2}}}\left( {{\text{e}}_1^{\text{2}} - 1} \right)\]
\[ \Rightarrow \dfrac{{{{\text{b}}^2}}}{{{{\text{a}}^{\text{2}}}}} = \left( {{\text{e}}_{\text{1}}^{\text{2}} - 1} \right)\]
Rearranging the terms,
\[ \Rightarrow {\text{e}}_{\text{1}}^{\text{2}} = \dfrac{{{{\text{b}}^2}}}{{{{\text{a}}^{\text{2}}}}} + 1\]
Hence,
$ \Rightarrow {{\text{e}}_{\text{1}}} = \sqrt {1 + \dfrac{{{{\text{b}}^{\text{2}}}}}{{{{\text{a}}^{\text{2}}}}}} $
If the transverse axis along the y-axis and the conjugate axis is along x-axis, then the equation of the hyperbola \[\dfrac{{{{\text{y}}^{\text{2}}}}}{{{{\text{b}}^{\text{2}}}}} - \dfrac{{{{\text{x}}^{\text{2}}}}}{{{{\text{a}}^{\text{2}}}}} = 1\] where \[{{\text{a}}^2} = {{\text{b}}^{\text{2}}}\left( {{\text{e}}_2^{\text{2}} - 1} \right)\] .
Let ${{\text{e}}_{\text{2}}}$ be the eccentricities of hyperbola \[\dfrac{{{{\text{y}}^{\text{2}}}}}{{{{\text{b}}^{\text{2}}}}} - \dfrac{{{{\text{x}}^{\text{2}}}}}{{{{\text{a}}^{\text{2}}}}} = 1\] where \[{{\text{a}}^2} = {{\text{b}}^{\text{2}}}\left( {{\text{e}}_2^{\text{2}} - 1} \right)\] .
Now, we are going to get ${{\text{e}}_{\text{2}}}$ from the term \[{{\text{a}}^2} = {{\text{b}}^{\text{2}}}\left( {{\text{e}}_2^{\text{2}} - 1} \right)\] .
\[ \Rightarrow \dfrac{{{{\text{a}}^2}}}{{{{\text{b}}^{\text{2}}}}} = \left( {{\text{e}}_2^{\text{2}} - 1} \right)\]
Rearranging the terms we get,
\[ \Rightarrow {\text{e}}_2^{\text{2}} = \dfrac{{{{\text{a}}^{\text{2}}}}}{{{{\text{b}}^{\text{2}}}}} + 1\]
Hence,
$ \Rightarrow {{\text{e}}_2} = \sqrt {1 + \dfrac{{{{\text{a}}^{\text{2}}}}}{{{{\text{b}}^{\text{2}}}}}} $
Let \[\dfrac{{{{\text{x}}^{\text{2}}}}}{{{{\text{a}}^{\text{2}}}}} - \dfrac{{{{\text{y}}^{\text{2}}}}}{{{{\text{b}}^{\text{2}}}}} = 1\] and \[\dfrac{{{{\text{y}}^{\text{2}}}}}{{{{\text{b}}^{\text{2}}}}} - \dfrac{{{{\text{x}}^{\text{2}}}}}{{{{\text{a}}^{\text{2}}}}} = 1\] be the two hyperbola conjugate to each other.
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