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Question Number 81698 by peter frank last updated on 14/Feb/20

Commented by mr W last updated on 14/Feb/20

question is wrong!

$${question}\:{is}\:{wrong}! \\ $$

Answered by mr W last updated on 14/Feb/20

y^2 =4x^2 ⇒y=±2x ⇒two lines y=mx+c=2x ⇒x_1 =(c/(2−m)), y_1 =((2c)/(2−m)) y=mx+c=−2x ⇒x_2 =−(c/(2+m)), y_2 =((2c)/(2+m)) distance=(√((x_2 −x_1 )^2 +(y_2 −y_1 )^2 )) =(√(((c/(2−m))+(c/(2+m)))^2 +(((2c)/(2−m))−((2c)/(2+m)))^2 )) =((4c(√(1+m^2 )))/(4−m^2 ))

$${y}^{\mathrm{2}} =\mathrm{4}{x}^{\mathrm{2}} \\ $$$$\Rightarrow{y}=\pm\mathrm{2}{x}\:\Rightarrow{two}\:{lines} \\ $$$${y}={mx}+{c}=\mathrm{2}{x} \\ $$$$\Rightarrow{x}_{\mathrm{1}} =\frac{{c}}{\mathrm{2}−{m}},\:{y}_{\mathrm{1}} =\frac{\mathrm{2}{c}}{\mathrm{2}−{m}} \\ $$$${y}={mx}+{c}=−\mathrm{2}{x} \\ $$$$\Rightarrow{x}_{\mathrm{2}} =−\frac{{c}}{\mathrm{2}+{m}},\:{y}_{\mathrm{2}} =\frac{\mathrm{2}{c}}{\mathrm{2}+{m}} \\ $$$$ \\ $$$${distance}=\sqrt{\left({x}_{\mathrm{2}} −{x}_{\mathrm{1}} \right)^{\mathrm{2}} +\left({y}_{\mathrm{2}} −{y}_{\mathrm{1}} \right)^{\mathrm{2}} } \\ $$$$=\sqrt{\left(\frac{{c}}{\mathrm{2}−{m}}+\frac{{c}}{\mathrm{2}+{m}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{2}{c}}{\mathrm{2}−{m}}−\frac{\mathrm{2}{c}}{\mathrm{2}+{m}}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{4}{c}\sqrt{\mathrm{1}+{m}^{\mathrm{2}} }}{\mathrm{4}−{m}^{\mathrm{2}} } \\ $$

Commented by peter frank last updated on 14/Feb/20

thank you sir.

$${thank}\:{you}\:{sir}. \\ $$

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