Question Number 51236 by peter frank last updated on 25/Dec/18
$${Prove}\:{that}\:{line}\:{y}={mx}+\frac{\mathrm{3}}{\mathrm{4}\:\:}{m}+\frac{\mathrm{1}}{{m}} \\ $$$${touches}\:{the}\:{parabola} \\ $$$${y}^{\mathrm{2}} =\mathrm{4}{x}+\mathrm{3}\:{whatever}\:{the} \\ $$$${value}\:{of}\:{m} \\ $$
Answered by mr W last updated on 25/Dec/18
$${y}={mx}+\frac{\mathrm{3}}{\mathrm{4}}{m}+\frac{\mathrm{1}}{{m}}\Rightarrow{x}=\frac{{y}}{{m}}−\frac{\mathrm{3}}{\mathrm{4}}−\frac{\mathrm{1}}{{m}^{\mathrm{2}} } \\ $$$${y}^{\mathrm{2}} =\mathrm{4}{x}+\mathrm{3}\Rightarrow{y}^{\mathrm{2}} =\frac{\mathrm{4}{y}}{{m}}−\mathrm{3}−\frac{\mathrm{4}}{{m}^{\mathrm{2}} }+\mathrm{3} \\ $$$$\Rightarrow{m}^{\mathrm{2}} {y}^{\mathrm{2}} −\mathrm{4}{my}+\mathrm{4}=\mathrm{0} \\ $$$$\Rightarrow\left({my}−\mathrm{2}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{y}=\frac{\mathrm{2}}{{m}} \\ $$$$\Rightarrow{x}=\frac{\mathrm{1}}{{m}^{\mathrm{2}} }−\frac{\mathrm{3}}{\mathrm{4}} \\ $$$${i}.{e}.\:{the}\:{line}\:{and}\:{the}\:{parabola}\:{intersect} \\ $$$${always}\:{at}\:{one}\:{and}\:{only}\:{one}\:{point} \\ $$$$\left(\frac{\mathrm{1}}{{m}^{\mathrm{2}} }−\frac{\mathrm{3}}{\mathrm{4}},\:\frac{\mathrm{2}}{{m}}\right)\:{for}\:{any}\:{value}\:{of}\:{m}\:{except} \\ $$$${zero}.\:{when}\:{two}\:{curves}\:{intersect}\:{at}\:{one} \\ $$$${and}\:{only}\:{one}\:{point},\:{that}\:{means}\:{they} \\ $$$${touch}\:{each}\:{other}. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 25/Dec/18
$$\:{y}={mx}+\frac{\mathrm{3}{m}}{\mathrm{4}}+\frac{\mathrm{1}}{{m}} \\ $$$${y}^{\mathrm{2}} =\mathrm{4}{x}+\mathrm{3} \\ $$$${solve}…{we}\:{have}\:{to}\:{prove}\:{roots}\:{are}\:{same} \\ $$$${x}_{\mathrm{1}} ={x}_{\mathrm{2}} \:\:\:{and}\:{y}_{\mathrm{1}} ={y}_{\mathrm{2}} \:\:{that}\:{means}\:{D}={B}^{\mathrm{2}} −\mathrm{4}{AC}=\mathrm{0} \\ $$$$\left({mx}+\frac{\mathrm{3}{m}}{\mathrm{4}}+\frac{\mathrm{1}}{{m}}\right)^{\mathrm{2}} =\mathrm{4}{x}+\mathrm{3} \\ $$$${m}^{\mathrm{2}} {x}^{\mathrm{2}} +\frac{\mathrm{9}{m}^{\mathrm{2}} }{\mathrm{16}}+\frac{\mathrm{1}}{{m}^{\mathrm{2}} }+\frac{\mathrm{6}{m}^{\mathrm{2}} {x}}{\mathrm{4}}+\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{2}{x}=\mathrm{4}{x}+\mathrm{3} \\ $$$${x}^{\mathrm{2}} \left({m}^{\mathrm{2}} \right)+{x}\left(\frac{\mathrm{6}{m}^{\mathrm{2}} }{\mathrm{4}}+\mathrm{2}−\mathrm{4}\right)+\left(\frac{\mathrm{9}{m}^{\mathrm{2}} }{\mathrm{16}}+\frac{\mathrm{1}}{{m}^{\mathrm{2}} }+\frac{\mathrm{3}}{\mathrm{2}}−\mathrm{3}\right)=\mathrm{0} \\ $$$${B}^{\mathrm{2}} =\left(\frac{\mathrm{3}{m}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{2}\right)^{\mathrm{2}} =\frac{\mathrm{9}{m}^{\mathrm{4}} }{\mathrm{4}}−\mathrm{2}×\frac{\mathrm{3}{m}^{\mathrm{2}} }{\mathrm{2}}×\mathrm{2}+\mathrm{4}=\frac{\mathrm{9}{m}^{\mathrm{4}} }{\mathrm{4}}−\mathrm{6}{m}^{\mathrm{2}} +\mathrm{4} \\ $$$$\mathrm{4}{AC}=\mathrm{4}×{m}^{\mathrm{2}} ×\left(\frac{\mathrm{9}{m}^{\mathrm{2}} }{\mathrm{16}}+\frac{\mathrm{1}}{{m}^{\mathrm{2}} }−\frac{\mathrm{3}}{\mathrm{2}}\right) \\ $$$$=\frac{\mathrm{9}{m}^{\mathrm{4}} }{\mathrm{4}}+\mathrm{4}−\mathrm{6}{m}^{\mathrm{2}} \\ $$$$\boldsymbol{{B}}^{\mathrm{2}} =\mathrm{4}{AC} \\ $$$${hence}\:{proved}… \\ $$$$ \\ $$
Commented by peter frank last updated on 25/Dec/18
$${thank}\:{you}\:{sir}.{what}\:{is}\:{value} \\ $$$${of}\:{m} \\ $$