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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 ( $\left.x _{1}, y _{1}\right)$ and $C _{2}$ at $\left( x _{2}, y _{2}\right)$, then $\left|2 x _{1}+ x _{2}\right|$ is equal to $......$
$19$
$18$
$17$
$20$
Solution
$T_{1}: y=m x \pm \sqrt{4 m^{2}+9}$
And $T_{2}: y=m x \pm \sqrt{42 m^{2}-13}$
So, $4\,m^{2}+9=42 m^{2}-143$
$38\,m ^{2}=152$
$m=\pm 2$
$c=\pm 5$
For given tangent not pass through $4^{\text {th }}$ quadrant
$T: y=2 x+5$
Now, comparing with $\frac{ xx _{1}}{4}+\frac{ yy _{1}}{9}=1$
We get, $\frac{x_{1}}{8}=-\frac{1}{5} \Rightarrow x_{1}=-\frac{8}{5}$
$\frac{ xx _{2}}{42}-\frac{ yy _{2}}{143}=1$
$2 x-y=-5$ we have
$x _{2}=-\frac{84}{5}$
So, $\left|2 x _{1}+ x _{2}\right|=\left|\frac{-100}{5}\right|=20$
