A common tangent $\mathrm{T}$ to the curves $\mathrm{C}_{1}: \frac{x^{2}}{4}+\frac{y^{2}}{9}=1$ and $C_{2}: \frac{x^{2}}{42}-\frac{y^{2}}{143}=1$ does not pass through the fourth quadrant. If $\mathrm{T}$ touches $\mathrm{C}_{1}$ at $\left(x_{1}, y_{1}\right)$ and $\mathrm{C}_{2}$ at $\left(x_{2}, y_{2}\right)$, then $\left|2 x_{1}+x_{2}\right|$ is equal to ______________.
Answer (integer)
20
Solution
<p>Equation of tangent to ellipse ${{{x^2}} \over 4} + {{{y^2}} \over 9} = 1$ and given slope m is : $y = mx + \sqrt {4{m^2} + 9}$ ..... (i)</p>
<p>For slope m equation of tangent to hyperbola is :</p>
<p>$y = mx + \sqrt {42{m^2} - 143}$ ....... (ii)</p>
<p>Tangents from (i) and (ii) are identical then</p>
<p>$4{m^2} + 9 = 42{m^2} - 143$</p>
<p>$\therefore$ $m = \, \pm \,2$ (+2 is not acceptable)</p>
<p>$\therefore$ $m = - 2$.</p>
<p>Hence, ${x_1} = {8 \over 5}$ and ${x_2} = {{84} \over 5}$</p>
<p>$\therefore$ $|2{x_1} + {x_2}| = \left| {{{16} \over 5} + {{84} \over 5}} \right| = 20$</p>
About this question
Subject: Mathematics · Chapter: Conic Sections · Topic: Parabola
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