Question

The eccentricity of the hyperbola $\dfrac{\surd1999}{3}(x^{2}-y^{2})=1$ is
(A) $\surd3$
(B) $\surd2$
(C) 2
(D) $2\surd2$

Answer 1

Hint: The ratio of the distance from the centre to one of the foci (c) and the distance from the centre to one of the vertices makes up the eccentricity of a hyperbola (a). 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 given equation of hyperbola is $\dfrac{\surd1999}{3}(x^{2}-y^{2})=1$
We can see that the values of a and b are the same in the given equation. This type of hyperbola is known as rectangular hyperbola.
The eccentricity of the 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, the eccentricity of the hyperbola $\dfrac{\surd1999}{3}(x^{2}-y^{2})=1$ is $\surd2$



Option ‘B’ is correct

Additional Information: 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. When the transverse and conjugate axes of a hyperbola are both the same length, the hyperbola is said to be rectangular.

Note: It should be remembered that the eccentricity of a rectangular hyperbola always comes out to be $\surd2$ As is well known, the distance from the centre to the foci (c) in a hyperbola is greater than or equal to the distance from the centre to one of the vertices. Thus, the eccentricity of a hyperbola is always greater than or equal to one.