A common tangent $T$ to the curves $C_{1}: \frac{x^{2}}{4}+\frac{y^{2}}{9}=1$ and $C_{2}: \frac{x^{2

Question

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

Professor PC Thomas Classes · Answer

(20) : Equation of tangent of curve $C_{1}$ is $y=m x \pm \sqrt{4 m^{2}+9}$\ Equation of tangent of curve $C_{2}$ is ;\ $y=m x \pm \sqrt{42 m^{2}-143}$\ Since both the tangents are same so;\ $\sqrt{4 m^{2}+9}=\sqrt{42 m^{2}-143}$\ $\Rightarrow 4 m^{2}+9=42 m^{2}-143$\ $\Rightarrow 38 m^{2}=152 \Rightarrow m= \pm 2$\ As given tangent does not pass through IV quadrant.\ So, $y=2 x+5$\ Now comparing $y=2 x+5$ with\ $\frac{x x_{1}}{4}+\frac{y y_{1}}{9}=1$, we get\ $\frac{x_{1}}{4}=-\frac{2}{5} \Rightarrow x_{1}=-8 / 5$ and similarly\ compare $y=2 x+5$ with $\frac{x x_{2}}{42}-\frac{y y_{2}}{143}=1$, we get $x_{2}=\frac{-84}{5}$\ So, $\left|2 x_{1}+x_{2}ight|=\left|\frac{-16}{5}-\frac{84}{5}ight|=\left|\frac{-100}{5}ight|=20$\