Question Number 31639 by Tinkutara last updated on 11/Mar/18
Answered by MJS last updated on 11/Mar/18
$${P}\in{par}:\:\begin{pmatrix}{{p}}\\{\frac{{p}^{\mathrm{2}} }{\mathrm{4}}−\frac{{p}}{\mathrm{2}}+\frac{\mathrm{5}}{\mathrm{4}}}\end{pmatrix} \\ $$$${y}'\left({p}\right)=\frac{{p}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\:\left(={k}\:\mathrm{of}\:\mathrm{tangent}\:\mathrm{in}\:\mathrm{P}\right) \\ $$$${k}\:\mathrm{of}\:\mathrm{normal}\:{n}\:\mathrm{in}\:\mathrm{P}:\:−\left(\frac{{p}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{−\mathrm{1}} = \\ $$$$=−\frac{\mathrm{2}}{{p}−\mathrm{1}} \\ $$$${y}=−\frac{\mathrm{2}}{{p}−\mathrm{1}}{x}+{d} \\ $$$$\mathrm{in}\:{P}: \\ $$$$\frac{{p}^{\mathrm{2}} }{\mathrm{4}}−\frac{{p}}{\mathrm{2}}+\frac{\mathrm{5}}{\mathrm{4}}=−\frac{\mathrm{2}{p}}{{p}−\mathrm{1}}+{d} \\ $$$${d}=\frac{{p}^{\mathrm{3}} −\mathrm{3}{p}^{\mathrm{2}} +\mathrm{15}{p}−\mathrm{5}}{\mathrm{4}\left({p}−\mathrm{1}\right)} \\ $$$${n}_{{P}} :\:{y}=−\frac{\mathrm{2}}{{p}−\mathrm{1}}{x}+\frac{{p}^{\mathrm{3}} −\mathrm{3}{p}^{\mathrm{2}} +\mathrm{15}{p}−\mathrm{5}}{\mathrm{4}\left({p}−\mathrm{1}\right)} \\ $$$$\mathrm{now}\:\mathrm{looking}\:\mathrm{for}\:\mathrm{Q}\in{par}\:\mathrm{with} \\ $$$${k}\left({n}_{{Q}} \right)={k}\left({t}_{{P}} \right) \\ $$$$−\frac{\mathrm{2}}{{q}−\mathrm{1}}=\frac{{p}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${q}=\mathrm{1}−\frac{\mathrm{4}}{{p}−\mathrm{1}} \\ $$$${n}_{{Q}} :\:{y}=−\frac{\mathrm{2}}{{q}−\mathrm{1}}{x}+\frac{{q}^{\mathrm{3}} −\mathrm{3}{q}^{\mathrm{2}} +\mathrm{15}{q}−\mathrm{5}}{\mathrm{4}\left({q}−\mathrm{1}\right)} \\ $$$$\Rightarrow \\ $$$${n}_{{Q}} :\:{y}=\frac{{p}−\mathrm{1}}{\mathrm{2}}{x}−\frac{{p}^{\mathrm{3}} −\mathrm{9}{p}^{\mathrm{2}} +\mathrm{15}{p}−\mathrm{15}}{\mathrm{2}\left({p}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${n}_{{P}} \cap{n}_{{Q}} \\ $$$${x}=\frac{{p}^{\mathrm{2}} −\mathrm{5}}{\mathrm{2}\left({p}−\mathrm{1}\right)};\:{y}=\frac{\left({p}^{\mathrm{2}} −\mathrm{2}{p}+\mathrm{5}\right)^{\mathrm{2}} }{\mathrm{4}\left({p}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\:{y}={x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{5} \\ $$$$\mathrm{points}\:\mathrm{of}\:\mathrm{intersection}\:\mathrm{lie}\:\mathrm{on} \\ $$$$\mathrm{a}\:\mathrm{parabola} \\ $$
Commented by Tinkutara last updated on 11/Mar/18
$${But}\:{answer}\:{is}\:\left({x}−\mathrm{1}\right)^{\mathrm{2}} ={y}−\mathrm{2} \\ $$$${Is}\:{it}\:{wrong}\:{in}\:{book}? \\ $$
Commented by MJS last updated on 11/Mar/18
$$...\mathrm{that}\:\mathrm{would}\:\mathrm{be} \\ $$$${y}={x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{3} \\ $$$$\mathrm{I}\:\mathrm{will}\:\mathrm{check}\:\mathrm{it}\:\mathrm{again} \\ $$
Commented by MJS last updated on 11/Mar/18
$$\mathrm{I}\:\mathrm{found}\:\mathrm{no}\:\mathrm{mistake}\:\mathrm{in}\:\mathrm{my}\:\mathrm{calculation}, \\ $$$$\mathrm{please}\:\mathrm{check}\:\mathrm{it}\:\mathrm{again} \\ $$
Commented by Tinkutara last updated on 12/Mar/18
$${I}\:{saw}\:{again}\:{but}\:{answer}\:{given}\:{is}\:{only} \\ $$$$\left({x}−\mathrm{1}\right)^{\mathrm{2}} ={y}−\mathrm{2}.\:{How}\:{to}\:{verify}\:{which}\:{is} \\ $$$${correct}? \\ $$
Commented by MJS last updated on 12/Mar/18
$$\mathrm{let}'\mathrm{s}\:\mathrm{check}\:\mathrm{for}\:\mathrm{one}\:\mathrm{pair}\:\mathrm{of}\:\mathrm{points} \\ $$$${f}\left({x}\right)=\frac{{x}^{\mathrm{2}} }{\mathrm{4}}−\frac{{x}}{\mathrm{2}}+\frac{\mathrm{5}}{\mathrm{4}} \\ $$$${f}'\left({x}\right)=\frac{{x}−\mathrm{1}}{\mathrm{2}}={k}_{{tangents}} ={k}_{{t}} \\ $$$${k}_{{normals}} =−\frac{\mathrm{2}}{{x}−\mathrm{1}}={k}_{{n}} \\ $$$${P}=\begin{pmatrix}{\mathrm{7}}\\{\mathrm{10}}\end{pmatrix};\:{k}_{{t}} =\mathrm{3};\:{k}_{{n}} =−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\mathrm{normal}\:\mathrm{in}\:{P}: \\ $$$${y}=−\frac{{x}}{\mathrm{3}}+{d}\:\Rightarrow\:\mathrm{10}=−\frac{\mathrm{7}}{\mathrm{3}}+{d}\:\Rightarrow\:{d}=\frac{\mathrm{37}}{\mathrm{3}} \\ $$$${n}_{{P}} :\:{y}=−\frac{{x}}{\mathrm{3}}+\frac{\mathrm{37}}{\mathrm{3}} \\ $$$$\mathrm{now}\:\mathrm{we}\:\mathrm{search}\:{Q}\:\mathrm{with}\:{Q}\in{par}\:\mathrm{and} \\ $$$${n}_{{Q}} \parallel{t}_{{P}} \:\left(\Rightarrow\:{n}_{{Q}} \bot{n}_{{P}} \right) \\ $$$${Q}=\begin{pmatrix}{{q}}\\{{f}\left({q}\right)}\end{pmatrix} \\ $$$${n}_{{Q}} :\:{y}={kx}+{d}\:\mathrm{with}\:{k}={k}_{{n}} =−\frac{\mathrm{2}}{{q}−\mathrm{1}} \\ $$$$\mathrm{and}\:{k}_{{n}} \:\mathrm{in}\:{Q}\:=\:{k}_{{t}} \:\mathrm{in}\:{P}\:\:\Rightarrow \\ $$$$\Rightarrow\:−\frac{\mathrm{2}}{{q}−\mathrm{1}}=\mathrm{3}\:\Rightarrow\:{q}=\frac{\mathrm{1}}{\mathrm{3}}\:\Rightarrow\:{f}\left({q}\right)=\frac{\mathrm{10}}{\mathrm{9}} \\ $$$${Q}=\begin{pmatrix}{\frac{\mathrm{1}}{\mathrm{3}}}\\{\frac{\mathrm{10}}{\mathrm{9}}}\end{pmatrix} \\ $$$$\mathrm{normal}\:\mathrm{in}\:{Q}: \\ $$$${y}=\mathrm{3}{x}+{d}\:\Rightarrow\:\frac{\mathrm{10}}{\mathrm{9}}=\mathrm{1}+{d}\:\Rightarrow\:{d}=\frac{\mathrm{1}}{\mathrm{9}} \\ $$$${n}_{{Q}} :\:{y}=\mathrm{3}{x}+\frac{\mathrm{1}}{\mathrm{9}} \\ $$$$\mathrm{intersection}\:\mathrm{of}\:{n}_{{P}} \:\mathrm{and}\:{n}_{{Q}} : \\ $$$${n}_{{P}} :\:{y}=−\frac{{x}}{\mathrm{3}}+\frac{\mathrm{37}}{\mathrm{3}} \\ $$$${n}_{{Q}} :\:{y}=\mathrm{3}{x}+\frac{\mathrm{1}}{\mathrm{9}} \\ $$$$\mathrm{3}{x}+\frac{\mathrm{1}}{\mathrm{9}}=−\frac{{x}}{\mathrm{3}}+\frac{\mathrm{37}}{\mathrm{3}}\:\Rightarrow\:{x}=\frac{\mathrm{11}}{\mathrm{3}}\:\Rightarrow\:{y}=\frac{\mathrm{100}}{\mathrm{9}} \\ $$$${S}_{{n}_{{P}} {n}_{{Q}} } =\begin{pmatrix}{\frac{\mathrm{11}}{\mathrm{3}}}\\{\frac{\mathrm{100}}{\mathrm{9}}}\end{pmatrix} \\ $$$$\mathrm{intersection}\:\mathrm{point}\:\mathrm{on}\:``\mathrm{my}''\:\mathrm{parabola}: \\ $$$${y}={x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{5}=\frac{\mathrm{100}}{\mathrm{9}}\: \\ $$$$\mathrm{or}\:\mathrm{on}\:``\mathrm{book}'\mathrm{s}''\:\mathrm{parabola}: \\ $$$${y}={x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{3}=\frac{\mathrm{82}}{\mathrm{9}} \\ $$$$\mathrm{so}\:\mathrm{the}\:\mathrm{book}\:\mathrm{is}\:\mathrm{wrong} \\ $$
Commented by Tinkutara last updated on 12/Mar/18
Thank you very much Sir! I got the answer. ��������