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

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Question

$T_{1}: y=m x \pm \sqrt{4 m^{2}+9}$

And $T_{2}: y=m x \pm \sqrt{42 m^{2}-13}$

So, $4\,m^{2}+9=42 m^{2}-143$

$38\,m ^{2}=152$

$m=\pm 2$

Vedclass · Accepted Answer

$20$

Vedclass · Answer

$T_{1}: y=m x \pm \sqrt{4 m^{2}+9}$

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 $4^{	ext {th }}$ 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, $\left|2 x _{1}+ x _{2}ight|=\left|\frac{-100}{5}ight|=20$

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