
If the transverse and conjugate axes of a hyperbola are equal, then its eccentricity is
(A)$ \surd3$
(B) $\surd2$
(C) $\dfrac{1}{\surd2}$
(D) 2
Answer
161.4k+ views
Hint: The set of all points in a plane whose difference in distance from two fixed points, referred to as foci, in the plane is a constant is known as a hyperbola. A hyperbola is referred to as a rectangular hyperbola if the length of the transverse axis (2a) in the hyperbola equals the length of the conjugate axis (2b).
Complete step by step solution:The equation of a hyperbola is given by $\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1$
When the transverse and conjugate axes of a hyperbola are equal, that is a = b. Then, the equation of hyperbola becomes$x^{2}-y^{2}=a^{2}$
It is a rectangular hyperbola.
The eccentricity of a rectangular hyperbola is given by e= $\dfrac{c}{a}.$
Now, $c^2=a^2+b^2$
As a = b, hence, $c^2=a^2+a^2=2a^2$
Thus, $c=\surd2a$
$e= \dfrac{c}{a}$ = $\dfrac{\surd2a}{a}$ = $\surd2$
Thus, If transverse and conjugate axes of a hyperbola are equal, then its eccentricity is $\surd2$.
Option ‘B’ is correct
Note: It should be remembered that the eccentricity of a rectangular hyperbola always comes out to be $\surd2$. In a hyperbola, there are two axes. The transverse axis is the line that passes through the foci. The conjugate axis is the line through the centre that is also perpendicular to the transverse axis.As we know, in the case of a hyperbola, the distance from the centre to the foci (c) is greater than or equal to the distance from the centre to one of the vertices. Therefore, the eccentricity of a hyperbola is always greater than or equal to one.
Complete step by step solution:The equation of a hyperbola is given by $\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1$
When the transverse and conjugate axes of a hyperbola are equal, that is a = b. Then, the equation of hyperbola becomes$x^{2}-y^{2}=a^{2}$
It is a rectangular hyperbola.
The eccentricity of a rectangular hyperbola is given by e= $\dfrac{c}{a}.$
Now, $c^2=a^2+b^2$
As a = b, hence, $c^2=a^2+a^2=2a^2$
Thus, $c=\surd2a$
$e= \dfrac{c}{a}$ = $\dfrac{\surd2a}{a}$ = $\surd2$
Thus, If transverse and conjugate axes of a hyperbola are equal, then its eccentricity is $\surd2$.
Option ‘B’ is correct
Note: It should be remembered that the eccentricity of a rectangular hyperbola always comes out to be $\surd2$. In a hyperbola, there are two axes. The transverse axis is the line that passes through the foci. The conjugate axis is the line through the centre that is also perpendicular to the transverse axis.As we know, in the case of a hyperbola, the distance from the centre to the foci (c) is greater than or equal to the distance from the centre to one of the vertices. Therefore, the eccentricity of a hyperbola is always greater than or equal to one.
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