Question

If \[{e_1}\] and \[{e_2}\] are the eccentricities of a hyperbola and its conjugate, then \[e_1^2 + e_2^2\] will be
(a) 1
(b) \[e_1^2e_2^2\]
(c) 0
(d) \[\dfrac{1}{{e_1^2}} + \dfrac{1}{{e_2^2}}\]

Answer 1

Hint:
Here, we need to find the value of \[e_1^2 + e_2^2\]. Using the equations of the hyperbola and its conjugate, we can find the eccentricities and simplify the expression \[e_1^2 + e_2^2\]. We can then compare the expression with the options given to find the correct option.
Formula Used: We will use the formulae eccentricity of hyperbola, \[e = \sqrt {1 + \dfrac{{{b^2}}}{{{a^2}}}} \], eccentricity of conjugate hyperbola, \[e = \sqrt {1 + \dfrac{{{a^2}}}{{{b^2}}}} \], where \[e\] is the eccentricity, \[a\] is the length of semi- major axes and \[b\] is the length of semi-minor axes.

Complete step by step solution:
We know that the equation of a hyperbola is given by \[\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1\].
The eccentricity of a hyperbola \[e\] is given by the formula \[e = \sqrt {1 + \dfrac{{{b^2}}}{{{a^2}}}} \].
Therefore, we can write \[{e_1}\] as
\[{e_1} = \sqrt {1 + \dfrac{{{b^2}}}{{{a^2}}}} \]
Squaring both sides, we get
\[ \Rightarrow e_1^2 = 1 + \dfrac{{{b^2}}}{{{a^2}}}\]
The equation of a conjugate hyperbola is given by \[\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = - 1\].
The eccentricity of a conjugate hyperbola \[e\] is given by the formula \[e = \sqrt {1 + \dfrac{{{a^2}}}{{{b^2}}}} \].
Therefore, we can write \[{e_2}\] as
\[{e_2} = \sqrt {1 + \dfrac{{{a^2}}}{{{b^2}}}} \]
Squaring both sides, we get
\[ \Rightarrow e_2^2 = 1 + \dfrac{{{a^2}}}{{{b^2}}}\]
Now, we will substitute the values of the squares of the eccentricities in the expression \[e_1^2 + e_2^2\].
Substituting \[1 + \dfrac{{{b^2}}}{{{a^2}}}\] for \[e_1^2\] and \[1 + \dfrac{{{a^2}}}{{{b^2}}}\] for \[e_2^2\] in the expression, we get
\[\begin{array}{l} \Rightarrow e_1^2 + e_2^2 = 1 + \dfrac{{{b^2}}}{{{a^2}}} + 1 + \dfrac{{{a^2}}}{{{b^2}}}\\ \Rightarrow e_1^2 + e_2^2 = \dfrac{{{a^2}}}{{{b^2}}} + \dfrac{{{b^2}}}{{{a^2}}} + 2\end{array}\]
Next, we will verify the given options one by one to find out the answer.
We know that \[{a^2}\] and \[{b^2}\] are greater than or equal to 0.
Thus, the expression \[e_1^2 + e_2^2 = \dfrac{{{a^2}}}{{{b^2}}} + \dfrac{{{b^2}}}{{{a^2}}} + 2\] is greater than 2.
Therefore, the expression cannot be equal to 1 or 0. Options (a) and (c) are incorrect.
Now, we will simplify the expression in option (b) and compare it to the value of the expression \[e_1^2 + e_2^2\].
Substituting \[1 + \dfrac{{{b^2}}}{{{a^2}}}\] for \[e_1^2\] and \[1 + \dfrac{{{a^2}}}{{{b^2}}}\] for \[e_2^2\] in the expression \[e_1^2e_2^2\], we get
\[e_1^2e_2^2 = \left( {1 + \dfrac{{{b^2}}}{{{a^2}}}} \right)\left( {1 + \dfrac{{{a^2}}}{{{b^2}}}} \right)\]
Using the distributive property of multiplication, we get
\[\begin{array}{l} \Rightarrow e_1^2e_2^2 = 1 + \dfrac{{{b^2}}}{{{a^2}}} + \dfrac{{{a^2}}}{{{b^2}}} + \dfrac{{{b^2}}}{{{a^2}}} \cdot \dfrac{{{a^2}}}{{{b^2}}}\\ \Rightarrow e_1^2e_2^2 = 1 + \dfrac{{{b^2}}}{{{a^2}}} + \dfrac{{{a^2}}}{{{b^2}}} + 1\\ \Rightarrow e_1^2e_2^2 = \dfrac{{{a^2}}}{{{b^2}}} + \dfrac{{{b^2}}}{{{a^2}}} + 2\end{array}\]
This is equal to the simplified value of the expression \[e_1^2 + e_2^2\].

\[\therefore\] We get the correct option as option (c).

Note:
Hyperbola is a conic section defined as a plain curve such that the distance between two points is always constant. For hyperbola, eccentricity is defined as the ratio of the distance from focus to the vertices. Eccentricity of the hyperbola is greater than 1.