A conic passing through origin has its foci at \[\left( {5,{\text{ }}12} \right)\]&\[\left( {24,{\text{ }}7} \right)\]. Then its eccentricity of a hyperbola is
A. $\dfrac{{\sqrt {386} }}{{12}}$
B. $\dfrac{{\sqrt {386} }}{{39}}$
C. $\dfrac{{\sqrt {386} }}{{47}}$
D. $\dfrac{{\sqrt {386} }}{{51}}$

Hint:To solve this question we will use the formula of calculating the distance between foci in the hyperbola $2ae=\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}-\text{ }{{({{y}_{2}}-{{y}_{1}})}^{2}}}$ and the formula of foci radii $2a = \sqrt {{x_1}^2 + {x_2}^2} - \sqrt {{y_1}^2 + {y_2}^2} $ . Using both the given foci of hyperbola we will first calculate the distance between the foci and consider it as the first equation and then find foci radii and consider it as the second equation. Then we will substitute the first equation in the second equation and derive the value of eccentricity.

Formula used
Foci radii : $2a = \sqrt {{x_1}^2 + {x_2}^2} - \sqrt {{y_1}^2 + {y_2}^2} $
Distance between foci of hyperbola : $2ae=\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}-\text{ }{{({{y}_{2}}-{{y}_{1}})}^{2}}}$

Complete step by step Solution:
First we will find the value of foci radii.
 Foci radii = $2a = \sqrt {{x_1}^2 + {x_2}^2} - \sqrt {{y_1}^2 + {y_2}^2} $
 $ = \sqrt {{{24}^2} + {7^2}} - \sqrt {{5^2} + {{12}^2}} $
  \[ = 12\]............(i)
Now we will find distance between foci of hyperbola,
Distance between foci = \[2ae\]=$2ae = \sqrt {{{({x_2} - {x_1})}^2} - {{({y_2} - {y_1})}^{^2}}} $
$2ae = \sqrt {{{(24 - 5)}^2} - {{(7 - 12)}^{^2}}} $
$2ae = \sqrt {{{(19)}^2} - {{( - 5)}^{^2}}} $
$2ae = \sqrt {386} $.......(ii)
We will now substitute the equation (i) in (ii).
$12e = \sqrt {386} $
$e = \dfrac{{\sqrt {386} }}{{12}}$

Therefore, the correct option is (A).

Note: The value of the eccentricity for a hyperbola is always greater than one while for a parabola it is equal to one and for an ellipse it is between zero and one. To solve these kind of questions we must remember the concept,properties and the formulas of every type of conic section.