Question Number 39470 by Raj Singh last updated on 06/Jul/18
Commented by maxmathsup by imad last updated on 07/Jul/18
$${x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \:=\mathrm{1}\:\Rightarrow{y}^{\mathrm{2}} =\mathrm{1}−{x}^{\mathrm{2}} \:\Rightarrow\mathrm{1}−{x}^{\mathrm{2}} \:=\mathrm{4}{x}−\mathrm{4}\:\Rightarrow{x}^{\mathrm{2}} \:+\mathrm{4}{x}\:−\mathrm{5}\:=\mathrm{0}\Rightarrow \\ $$$$\Delta^{'} =\mathrm{2}^{\mathrm{2}} \:\:+\mathrm{5}\:=\mathrm{9}\:\:\Rightarrow\:{x}_{\mathrm{1}} =−\mathrm{2}\:+\mathrm{3}\:=\mathrm{1}\:\:{and}\:{x}_{\mathrm{2}} =−\mathrm{2}−\mathrm{3}\:=−\mathrm{5} \\ $$$${x}=\mathrm{1}\Rightarrow{y}=\mathrm{0}\:\Rightarrow\:{A}\left(\mathrm{1},\mathrm{0}\right)\:{is}\:{intersection} \\ $$$${x}=−\mathrm{5}\:\Rightarrow{y}^{\mathrm{2}} \:=−\mathrm{24}\:\:{and}\:{this}\:{is}\:{impossibleso}\:{the}\:{first}\:{point}\:{is}\:{A}. \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 07/Jul/18
Answered by ajfour last updated on 06/Jul/18
$${x}^{\mathrm{2}} +\mathrm{4}\left({x}−\mathrm{1}\right)=\mathrm{1} \\ $$$${x}^{\mathrm{2}} +\mathrm{4}{x}−\mathrm{5}=\mathrm{0} \\ $$$$\left({x}+\mathrm{2}\right)^{\mathrm{2}} =\mathrm{9} \\ $$$${x}=−\mathrm{2}\pm\mathrm{3}\:;\:\:{but}\:{for}\:{parabola}\:{x}\:\geqslant\:\mathrm{1} \\ $$$${hence}\:{only}\:{point}\:{is}\:\left(\mathrm{1},\mathrm{0}\right)\:;\:{hence} \\ $$$${they}\:{touch}\:\left({circle}\:{being}\:{a}\:{closed}\right. \\ $$$$\left.{curve}\right). \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 07/Jul/18
$${solve}... \\ $$$${x}^{\mathrm{2}} +\mathrm{4}{x}−\mathrm{4}=\mathrm{1} \\ $$$${x}^{\mathrm{2}} +\mathrm{5}{x}−{x}−\mathrm{5}=\mathrm{0} \\ $$$${x}\left({x}+\mathrm{5}\right)−\mathrm{1}\left({x}+\mathrm{5}\right)=\mathrm{0} \\ $$$$\left({x}+\mathrm{5}\right)\left({x}−\mathrm{1}\right)=\mathrm{0} \\ $$$${x}=\mathrm{1},−\mathrm{5} \\ $$$${when}\:{x}=\mathrm{1}\:\:{y}=\mathrm{0} \\ $$$$\mathrm{25}+{y}^{\mathrm{2}} =\mathrm{1}\:\:{then}\:{y}^{\mathrm{2}} =−\mathrm{24}\:\:{which}\:{is}\:{not}\:{possible}. \\ $$$${a}\:{circle}\:{and}\:{parabola}\:{intesect}\:{at}\:{two}\:{distinct} \\ $$$${points}\:\boldsymbol{{B}}{ut}\:{here}\:{only}\:{one}\:{solution}\left(\mathrm{1},\mathrm{0}\right) \\ $$$${satisfy}\:{both}\:{equation}... \\ $$$${so}\:{thiz}\:{two}\:{curve}\:{circle}\:{and}\:{parabola}\:{can}\:{not} \\ $$$${intersect}\:{at}\:{single}\:{point}...\boldsymbol{{H}}{ence}\:{they}\:{touch} \\ $$$${each}\:{other}\:{at}\:\left(\mathrm{1},\mathrm{0}\right) \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 07/Jul/18
$${when}\:{touch}\:\:{each}\:{other}\:{slope}\:{value}\:{caculated}\:{st}\:{that}\:{with}\: \\ $$$${respect}\:{to}\:{both}\:{curve}\:{should}\:{be}\:{same} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{1} \\ $$$$\mathrm{2}{x}\:+\mathrm{2}{y}.\frac{{dy}}{{dx}}=\mathrm{0} \\ $$$$\frac{{dy}}{{dx}}\:{at}\left(\mathrm{1},\mathrm{0}\right)\:{is}\:\:{m}_{\mathrm{1}} =\left(\frac{−{x}}{{y}}\right)_{\mathrm{1}'\mathrm{0}} =\infty \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$${y}^{\mathrm{2}} =\mathrm{4}\left({x}−\mathrm{1}\right) \\ $$$$\mathrm{2}{y}\frac{{dy}}{{dx}}=\mathrm{4}\:\:\:\:\frac{{dy}}{{dx}}=\frac{\mathrm{2}}{{y}}\:\:{so}\:{slope}\:{at}\:\left(\mathrm{1},\mathrm{0}\right)\:\:{is}=\infty= \\ $$