A common tangent T to the curves ${\mathrm{C}}_1:\frac{{\mathrm{x}}^2}{4}+\frac{{\mathrm{y}}^2}{9}=1\text{ and }{\mathrm{C}}_2:\frac{{\mathrm{x}}^2}{42}-\frac{{\mathrm{y}}^2}{143}=1$ does not pass through the fourth quadrant. If T touches C₁ at $\left({\mathrm{x}}_1, {\mathrm{y}}_1\right)\text{ and }{\mathrm{C}}_2\text{ at }\left({\mathrm{x}}_2, {\mathrm{y}}_2\right)\text{, then }\left|2{\mathrm{x}}_1+{\mathrm{x}}_2\right|$ is equal to _________.

Let common tangents are

${\mathrm{T}}_1 : \mathrm{y}=\mathrm{mx}\pm \sqrt{4{\mathrm{m}}^2+9}$

& ${\mathrm{T}}_2 : \mathrm{y}=\mathrm{mx}\pm \sqrt{42{\mathrm{m}}^2-143}$

So, $4{\mathrm{m}}^2+9=42{\mathrm{m}}^2-143$

$$\begin{gathered} \Rightarrow 38{\mathrm{m}}^2=152 \\ \Rightarrow \mathrm{m}=\pm 2 \\ \& \mathrm{c}=\pm 5 \end{gathered}$$

For given tangent not pass through 4^th quadrant $\mathrm{T} : \mathrm{y}=2\mathrm{x}+5$

Now, comparing with $\frac{{\mathrm{xx}}_1}{4}+\frac{{\mathrm{yy}}_1}{9}=1$

We get, $\frac{{\mathrm{x}}_1}{8}=-\frac{1}{5} \Rightarrow {\mathrm{x}}_1=-\frac{8}{5}$

$\frac{{\mathrm{xx}}_2}{42}-\frac{{\mathrm{yy}}_2}{143}=1$

2x – y = –5 we have

${\mathrm{x}}_2=-\frac{84}{5}$

So, $\left|2{\mathrm{x}}_1+{\mathrm{x}}_2\right|=\left|\frac{-100}{5}\right|=20$