A common tangent T to the curves C_{1}: frac{ | vedclass

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(D) The equation of a tangent to $C_{1}: \frac{x^{2}}{4} + \frac{y^{2}}{9} = 1$ is $y = mx \pm \sqrt{4m^{2} + 9}$.The equation of a tangent to $C_{2}:

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$20$

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(D) The equation of a tangent to $C_{1}: \frac{x^{2}}{4} + \frac{y^{2}}{9} = 1$ is $y = mx \pm \sqrt{4m^{2} + 9}$.The equation of a tangent to $C_{2}: \frac{x^{2}}{42} - \frac{y^{2}}{143} = 1$ is $y = mx \pm \sqrt{42m^{2} - 143}$.For a common tangent,$4m^{2} + 9 = 42m^{2} - 143$.$38m^{2} = 152$ $\Rightarrow m^{2} = 4$ $\Rightarrow m = \pm 2$.For $m = 2$,the constant term $c^{2} = 4(4) + 9 = 25$,so $c = \pm 5$.The tangent $T$ does not pass through the fourth quadrant,so we choose $y = 2x + 5$.The point of contact $(x_{1}, y_{1})$ on $C_{1}$ is given by $\frac{xx_{1}}{4} + \frac{yy_{1}}{9} = 1$. Comparing $2x - y = -5$ with $\frac{x_{1}}{4}x + \frac{y_{1}}{9}y = 1$,we get $\frac{x_{1}/4}{2} = \frac{y_{1}/9}{-1} = \frac{1}{-5}$.Thus,$x_{1} = -8/5$.The point of contact $(x_{2}, y_{2})$ on $C_{2}$ is given by $\frac{xx_{2}}{42} - \frac{yy_{2}}{143} = 1$. Comparing $2x - y = -5$ with $\frac{x_{2}}{42}x - \frac{y_{2}}{143}y = 1$,we get $\frac{x_{2}/42}{2} = \frac{-y_{2}/143}{-1} = \frac{1}{-5}$.Thus,$x_{2} = -84/5$.Finally,$|2x_{1} + x_{2}| = |2(-8/5) - 84/5| = |-16/5 - 84/5| = |-100/5| = 20$.

$A$ 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 $|2x_{1} + x_{2}|$ is equal to $......$

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