A common tangent T to the curves C _ 1 : x ^ 2 4 + y ^ 2 9 = 1 and...

Given curve ${C}_{1}: \dfrac{{x}^{2}}{4} + \dfrac{{y}^{2}}{9} = 1$ and ${C}_{2}: \dfrac{{x}^{2}}{42} - \dfrac{{y}^{2}}{143} = 1$

Now let common tangents to the curves are

${T}_{1}: y = mx \pm \sqrt{4{m}^{2} + 9}$

and ${T}_{2}: y = mx \pm \sqrt{42{m}^{2} - 143}$

Since they both are same,

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

$⇒ 38{m}^{2} = 152$

$⇒ m = \pm 2$

So, $c = \pm \sqrt{4{m}^{2} + 9} = \pm 5$

For given tangent not pass through ${4}^{th}$ quadrant

So, $T: y = 2x + 5$

Now, comparing with point form of tangent $\dfrac{x{x}_{1}}{4} + \dfrac{y{y}_{1}}{9} = 1$

We get, $\dfrac{{x}_{1}}{8} = - \dfrac{1}{5} ⇒ {x}_{1} = - \dfrac{8}{5}$

And point form of tangent equation of hyperbola $\dfrac{x{x}_{2}}{42} - \dfrac{y{y}_{2}}{143} = 1 and 2x - y = - 5$

We get ${x}_{2} = - \dfrac{84}{5}$

So \left|2{x}_{1} + {x}_{2}\right| = \left|\dfrac{- 100}{5}\right| = 20