Question

The eccentricity of the hyperbola can never be equal to-
A. $\sqrt {\dfrac{9}{5}}$
B. $2\sqrt {\dfrac{1}{9}}$
C. $3\sqrt {\dfrac{1}{8}}$
D. $\sqrt 2$

Answer 1

Hint To solve this question, we need to know the basic theory related to the chapter conic section. As we know the eccentricity of the hyperbola is greater than 1. Therefore, in this question, to get the correct answer, we have to proceed by going through all the options. And if given eccentricity is violating the above condition that means eccentricity of the hyperbola can never be equal to that option.

Complete step-by-step solution:
We know the eccentricity of the hyperbola is greater than 1.
So,
First option-
$\sqrt {\dfrac{9}{5}}$= $\dfrac{3}{{\sqrt 5 }}$ >1
Second option-
$2\sqrt {\dfrac{1}{9}}$= $2 \times \dfrac{1}{3}$= $\dfrac{2}{3}$<1
Third option-
$3\sqrt {\dfrac{1}{8}}$= $\dfrac{3}{{2\sqrt 2 }}$ >1
Fourth option-
$\sqrt 2$= 1.414 >1
Thus, the eccentricity of the hyperbola can never be equal to $2\sqrt {\dfrac{1}{9}}$.
Therefore, option (B) is the correct answer.

Note As we know eccentricity means a measure of how much the deviation of the curve has occurred from the circularity of the given shape. Therefore, the eccentricity of the circle is equal 0, i.e. e = 0 but in the case of hyperbola we know, it is defined as the set of all points in a plane in which the difference of whose distances from two fixed points is constant. In other words, the distance from the fixed point in a plane bears a constant ratio greater than the distance from the fixed-line in a plane. Therefore, the eccentricity of the hyperbola is greater than 1, i.e. e > 1