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Question Number 51236 by peter frank last updated on 25/Dec/18

Prove that line y=mx+(3/(4  ))m+(1/m)  touches the parabola  y^2 =4x+3 whatever the  value of m

$${Prove}\:{that}\:{line}\:{y}={mx}+\frac{\mathrm{3}}{\mathrm{4}\:\:}{m}+\frac{\mathrm{1}}{{m}} \\ $$$${touches}\:{the}\:{parabola} \\ $$$${y}^{\mathrm{2}} =\mathrm{4}{x}+\mathrm{3}\:{whatever}\:{the} \\ $$$${value}\:{of}\:{m} \\ $$

Answered by mr W last updated on 25/Dec/18

y=mx+(3/4)m+(1/m)⇒x=(y/m)−(3/4)−(1/m^2 )  y^2 =4x+3⇒y^2 =((4y)/m)−3−(4/m^2 )+3  ⇒m^2 y^2 −4my+4=0  ⇒(my−2)^2 =0  ⇒y=(2/m)  ⇒x=(1/m^2 )−(3/4)  i.e. the line and the parabola intersect  always at one and only one point  ((1/m^2 )−(3/4), (2/m)) for any value of m except  zero. when two curves intersect at one  and only one point, that means they  touch each other.

$${y}={mx}+\frac{\mathrm{3}}{\mathrm{4}}{m}+\frac{\mathrm{1}}{{m}}\Rightarrow{x}=\frac{{y}}{{m}}−\frac{\mathrm{3}}{\mathrm{4}}−\frac{\mathrm{1}}{{m}^{\mathrm{2}} } \\ $$$${y}^{\mathrm{2}} =\mathrm{4}{x}+\mathrm{3}\Rightarrow{y}^{\mathrm{2}} =\frac{\mathrm{4}{y}}{{m}}−\mathrm{3}−\frac{\mathrm{4}}{{m}^{\mathrm{2}} }+\mathrm{3} \\ $$$$\Rightarrow{m}^{\mathrm{2}} {y}^{\mathrm{2}} −\mathrm{4}{my}+\mathrm{4}=\mathrm{0} \\ $$$$\Rightarrow\left({my}−\mathrm{2}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{y}=\frac{\mathrm{2}}{{m}} \\ $$$$\Rightarrow{x}=\frac{\mathrm{1}}{{m}^{\mathrm{2}} }−\frac{\mathrm{3}}{\mathrm{4}} \\ $$$${i}.{e}.\:{the}\:{line}\:{and}\:{the}\:{parabola}\:{intersect} \\ $$$${always}\:{at}\:{one}\:{and}\:{only}\:{one}\:{point} \\ $$$$\left(\frac{\mathrm{1}}{{m}^{\mathrm{2}} }−\frac{\mathrm{3}}{\mathrm{4}},\:\frac{\mathrm{2}}{{m}}\right)\:{for}\:{any}\:{value}\:{of}\:{m}\:{except} \\ $$$${zero}.\:{when}\:{two}\:{curves}\:{intersect}\:{at}\:{one} \\ $$$${and}\:{only}\:{one}\:{point},\:{that}\:{means}\:{they} \\ $$$${touch}\:{each}\:{other}. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 25/Dec/18

 y=mx+((3m)/4)+(1/m)  y^2 =4x+3  solve...we have to prove roots are same  x_1 =x_2    and y_1 =y_2   that means D=B^2 −4AC=0  (mx+((3m)/4)+(1/m))^2 =4x+3  m^2 x^2 +((9m^2 )/(16))+(1/m^2 )+((6m^2 x)/4)+(3/2)+2x=4x+3  x^2 (m^2 )+x(((6m^2 )/4)+2−4)+(((9m^2 )/(16))+(1/m^2 )+(3/2)−3)=0  B^2 =(((3m^2 )/2)−2)^2 =((9m^4 )/4)−2×((3m^2 )/2)×2+4=((9m^4 )/4)−6m^2 +4  4AC=4×m^2 ×(((9m^2 )/(16))+(1/m^2 )−(3/2))  =((9m^4 )/4)+4−6m^2   B^2 =4AC  hence proved...

$$\:{y}={mx}+\frac{\mathrm{3}{m}}{\mathrm{4}}+\frac{\mathrm{1}}{{m}} \\ $$$${y}^{\mathrm{2}} =\mathrm{4}{x}+\mathrm{3} \\ $$$${solve}...{we}\:{have}\:{to}\:{prove}\:{roots}\:{are}\:{same} \\ $$$${x}_{\mathrm{1}} ={x}_{\mathrm{2}} \:\:\:{and}\:{y}_{\mathrm{1}} ={y}_{\mathrm{2}} \:\:{that}\:{means}\:{D}={B}^{\mathrm{2}} −\mathrm{4}{AC}=\mathrm{0} \\ $$$$\left({mx}+\frac{\mathrm{3}{m}}{\mathrm{4}}+\frac{\mathrm{1}}{{m}}\right)^{\mathrm{2}} =\mathrm{4}{x}+\mathrm{3} \\ $$$${m}^{\mathrm{2}} {x}^{\mathrm{2}} +\frac{\mathrm{9}{m}^{\mathrm{2}} }{\mathrm{16}}+\frac{\mathrm{1}}{{m}^{\mathrm{2}} }+\frac{\mathrm{6}{m}^{\mathrm{2}} {x}}{\mathrm{4}}+\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{2}{x}=\mathrm{4}{x}+\mathrm{3} \\ $$$${x}^{\mathrm{2}} \left({m}^{\mathrm{2}} \right)+{x}\left(\frac{\mathrm{6}{m}^{\mathrm{2}} }{\mathrm{4}}+\mathrm{2}−\mathrm{4}\right)+\left(\frac{\mathrm{9}{m}^{\mathrm{2}} }{\mathrm{16}}+\frac{\mathrm{1}}{{m}^{\mathrm{2}} }+\frac{\mathrm{3}}{\mathrm{2}}−\mathrm{3}\right)=\mathrm{0} \\ $$$${B}^{\mathrm{2}} =\left(\frac{\mathrm{3}{m}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{2}\right)^{\mathrm{2}} =\frac{\mathrm{9}{m}^{\mathrm{4}} }{\mathrm{4}}−\mathrm{2}×\frac{\mathrm{3}{m}^{\mathrm{2}} }{\mathrm{2}}×\mathrm{2}+\mathrm{4}=\frac{\mathrm{9}{m}^{\mathrm{4}} }{\mathrm{4}}−\mathrm{6}{m}^{\mathrm{2}} +\mathrm{4} \\ $$$$\mathrm{4}{AC}=\mathrm{4}×{m}^{\mathrm{2}} ×\left(\frac{\mathrm{9}{m}^{\mathrm{2}} }{\mathrm{16}}+\frac{\mathrm{1}}{{m}^{\mathrm{2}} }−\frac{\mathrm{3}}{\mathrm{2}}\right) \\ $$$$=\frac{\mathrm{9}{m}^{\mathrm{4}} }{\mathrm{4}}+\mathrm{4}−\mathrm{6}{m}^{\mathrm{2}} \\ $$$$\boldsymbol{{B}}^{\mathrm{2}} =\mathrm{4}{AC} \\ $$$${hence}\:{proved}... \\ $$$$ \\ $$

Commented by peter frank last updated on 25/Dec/18

thank you sir.what is value  of m

$${thank}\:{you}\:{sir}.{what}\:{is}\:{value} \\ $$$${of}\:{m} \\ $$

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