Question

: If $e$ and $e'$ be the eccentricities of a hyperbola and its conjugate, then $\dfrac{1}{{{(e)}^{2}}}+\dfrac{1}{{{(e')}^{2}}}$ is equal to
A. $0$
B. $1$
C. $2$
D. None of these

Answer 1

Hint: Write the equation of hyperbola and its conjugate. Then find the eccentricities of both and substitute in $\dfrac{1}{{{(e)}^{2}}}+\dfrac{1}{{{(e')}^{2}}}$. You will get the answer.

Complete step-by-step answer:
In mathematics, a hyperbola is a type of smooth curve lying in a plane, defined by its geometric properties or by equations for which it is the solution set. A hyperbola has two pieces, called connected components or branches, that are mirror images of each other and resemble two infinite bows. The hyperbola is one of the three kinds of conic sections, formed by the intersection of a plane and a double cone. (The other conic sections are the parabola and the ellipse. A circle is a special case of an ellipse.) If the plane intersects both halves of the double cone but does not pass through the apex of the cones, then the conic is a hyperbola.
Each branch of the hyperbola has two arms which become straighter (lower curvature) further out from the center of the hyperbola. Diagonally opposite arms, one from each branch, tend in the limit to a common line, called the asymptote of those two arms. So there are two asymptotes, whose intersection is at the center of symmetry of the hyperbola, which can be thought of as the mirror point about which each branch reflects to form the other branch. In the case of the curve, $y(x)=\dfrac{1}{x}$the asymptotes are the two coordinate axes.
The canonical form of a hyperbola is $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$.
For a hyperbola in the above canonical form, the eccentricity is given by $e=\sqrt{1+\dfrac{{{b}^{2}}}{{{a}^{2}}}}$.
So, the equation of the hyperbola is $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$.
Let $e$ be the eccentricity of the above hyperbola.
Since ${{b}^{2}}={{a}^{2}}({{e}^{2}}-1)$.
Simplifying we get,

\[{{e}^{2}}=1+\dfrac{{{b}^{2}}}{{{a}^{2}}}=\dfrac{{{a}^{2}}+{{b}^{2}}}{{{a}^{2}}}\]…………… (1)

So, the equation of conjugate of the above hyperbola is $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=-1$.
Let $e'$be the eccentricity of the above hyperbola.
Since ${{a}^{2}}={{b}^{2}}({{(e')}^{2}}-1)$.
Simplifying we get,

\[{{(e')}^{2}}=1+\dfrac{{{a}^{2}}}{{{b}^{2}}}=\dfrac{{{b}^{2}}+{{a}^{2}}}{{{b}^{2}}}\]…………… (2)

Now we want to find $\dfrac{1}{{{(e)}^{2}}}+\dfrac{1}{{{(e')}^{2}}}$.
So now substituting (1) and (2) in the above equation we get,
\[\dfrac{1}{{{(e)}^{2}}}+\dfrac{1}{{{(e')}^{2}}}=\dfrac{1}{\dfrac{{{a}^{2}}+{{b}^{2}}}{{{a}^{2}}}}+\dfrac{1}{\dfrac{{{b}^{2}}+{{a}^{2}}}{{{b}^{2}}}}\]
Simplifying we get,
\[\dfrac{1}{{{(e)}^{2}}}+\dfrac{1}{{{(e')}^{2}}}=\dfrac{{{a}^{2}}}{{{a}^{2}}+{{b}^{2}}}+\dfrac{{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}=\dfrac{{{a}^{2}}+{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}=1\]

Here, we had got the value of\[\dfrac{1}{{{(e)}^{2}}}+\dfrac{1}{{{(e')}^{2}}}\] which is $1$.

So, the correct answer is an option (B).

Note: Read the question carefully. You should know the equation of the eccentricity of the hyperbola. Also, many students make mistakes in writing the conjugate of the hyperbola. So avoid the mistake. Generally, the mistakes are in signs, so avoid it.