A common tangent T to the curves $C_{1} : \frac{x^{2}}{4} + \frac{y^{2}}{9} = 1$ and $C_{2} : \frac{x^{2}}{42} - \frac{y^{2}}{143} = 1$ does not pass through the fourth quadrant. If T touches $C_{1}$ at $(x_{1}, y_{1})$ and $C_{2}$ at $(x_{2}, y_{2}),$ then $| 2 x_{1} + x_{2} |$ is equal to_____

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answer is B.

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Detailed Solution

Let common tangents are $T_{1} : y = m x \pm \sqrt{4 m^{2} + 9}$
And $T_{2} : y = m x \pm \sqrt{42 m^{2} - 143}$
So, $4 m^{2} + 9 = 42 m^{2} - 143$
$\Rightarrow 38 m^{2} = 152 \Rightarrow m = \pm 2$ and $c = \pm 5$
For given tangent not pass through 4th quadrant
$T : y = 2 x + 5$
Now, comparing with $\frac{x x_{1}}{4} + \frac{y y_{1}}{9} = 1$
We get, $\frac{x_{1}}{8} = - \frac{1}{5} \Rightarrow x_{1} = - \frac{8}{5}; \frac{x x_{2}}{42} - \frac{y y_{2}}{143} = 1$
$2 x - y = - 5,$ Therefore, $x_{2} = - \frac{84}{5}$
Thus, $| 2 x_{1} + x_{2} | = | \frac{- 100}{5} | = 20$