Question Number 89415 by ajfour last updated on 17/Apr/20
Commented by ajfour last updated on 17/Apr/20
$${Ellipse}:\:\:\frac{{x}^{\mathrm{2}} }{\mathrm{4}}+{y}^{\mathrm{2}} =\mathrm{1} \\ $$$${And}\:{if}\:\:\angle{OP}\:{A}\:=\:\mathrm{90}°\:,\:{Find}\:{eq}. \\ $$$${of}\:{shown}\:{parabola}. \\ $$
Commented by Tony Lin last updated on 17/Apr/20
$${let}\:{P}\left(\mathrm{2}{cos}\theta,{sin}\theta\right) \\ $$$$\left(\mathrm{2}{cos}\theta,{sin}\theta\right)\centerdot\left(\mathrm{2}{cos}\theta−\mathrm{2},{sin}\theta\right)=\mathrm{0} \\ $$$$\mathrm{3}{cos}^{\mathrm{2}} \theta−\mathrm{4}{cos}\theta+\mathrm{1}=\mathrm{0} \\ $$$$\left(\mathrm{3}{cos}\theta−\mathrm{1}\right)\left({cos}\theta−\mathrm{1}\right)=\mathrm{0} \\ $$$${cos}\theta=\frac{\mathrm{1}}{\mathrm{3}}\:{or}\:\mathrm{1}\left({false}\right) \\ $$$$\Rightarrow{P}\left(\frac{\mathrm{2}}{\mathrm{3}},\:\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}}\right) \\ $$$${let}\:{parabola}\:{y}={a}\left({x}−{k}\right)^{\mathrm{2}} \\ $$$$\begin{cases}{\mathrm{1}={ak}^{\mathrm{2}} }\\{\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}}={a}\left(\frac{\mathrm{2}}{\mathrm{3}}−{k}\right)^{\mathrm{2}} }\end{cases} \\ $$
Commented by ajfour last updated on 17/Apr/20
$${this}\:{cannot}\:{be}\:{eq}.\:{of}\:{the}\:{required} \\ $$$${parabola},\:{please}\:{solve}\:{again},\:{Sir}. \\ $$
Commented by Tony Lin last updated on 17/Apr/20
$$\Rightarrow\begin{cases}{{a}=\frac{\mathrm{9}}{\mathrm{4}}+\frac{\mathrm{3}}{\:\sqrt{\mathrm{2}}}−\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\:\sqrt[{\mathrm{4}}]{\mathrm{2}}}}\\{{k}=\left(\mathrm{2}+\frac{\mathrm{4}\sqrt{\mathrm{2}}}{\mathrm{3}}\right)\left(\mathrm{3}+\sqrt{\mathrm{6}\sqrt{\mathrm{2}}}\right)}\end{cases} \\ $$$$\Rightarrow{y}=\left(\frac{\mathrm{9}}{\mathrm{4}}+\frac{\mathrm{3}}{\:\sqrt{\mathrm{2}}}−\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\:\sqrt[{\mathrm{4}}]{\mathrm{2}}}\right)\left\{{x}−\left(\mathrm{2}+\frac{\mathrm{4}\sqrt{\mathrm{2}}}{\mathrm{3}}\right)\left(\mathrm{3}+\sqrt{\mathrm{6}\sqrt{\mathrm{2}}}\right)\right\}^{\mathrm{2}} \\ $$
Answered by mr W last updated on 18/Apr/20
$$\frac{{x}^{\mathrm{2}} }{\mathrm{4}}+{y}^{\mathrm{2}} =\mathrm{1} \\ $$$${P}\left(\mathrm{2}\:\mathrm{cos}\:\theta,\:\mathrm{sin}\:\theta\right) \\ $$$$\left(\mathrm{2}\:\mathrm{cos}\:\theta\right)^{\mathrm{2}} +\mathrm{sin}^{\mathrm{2}} \:\theta+\mathrm{4}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)^{\mathrm{2}} +\mathrm{sin}^{\mathrm{2}} \:\theta=\mathrm{4} \\ $$$$\mathrm{3}\:\mathrm{cos}^{\mathrm{2}} \:\theta−\mathrm{4}\:\mathrm{cos}\:\theta+\mathrm{1}=\mathrm{0} \\ $$$$\left(\mathrm{3}\:\mathrm{cos}\:\theta−\mathrm{1}\right)\left(\mathrm{cos}\:\theta−\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{cos}\:\theta=\frac{\mathrm{1}}{\mathrm{3}}\:\Rightarrow\mathrm{sin}\:\theta=\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}} \\ $$$${P}\left(\frac{\mathrm{2}}{\mathrm{3}},\:\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}}\right) \\ $$$${eqn}.\:{of}\:{parabola}:\:{say}\:{y}={A}\left({x}−{B}\right)^{\mathrm{2}} \\ $$$$\mathrm{1}={AB}^{\mathrm{2}} \:\:\:…\left({i}\right) \\ $$$$\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}}={A}\left(\frac{\mathrm{2}}{\mathrm{3}}−{B}\right)^{\mathrm{2}} \:\:\:…\left({ii}\right) \\ $$$$\left(\frac{\mathrm{2}}{\mathrm{3}{B}}−\mathrm{1}\right)^{\mathrm{2}} =\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}} \\ $$$${B}=\frac{\mathrm{2}}{\mathrm{3}\pm\sqrt{\mathrm{6}\sqrt{\mathrm{2}}}} \\ $$$$\Rightarrow{B}=\frac{\mathrm{2}}{\mathrm{3}+\sqrt{\mathrm{6}\sqrt{\mathrm{2}}}}<\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${A}=\frac{\mathrm{1}}{{B}^{\mathrm{2}} } \\ $$$$ \\ $$$${eqn}.\:: \\ $$$${y}=\frac{\mathrm{1}}{{B}^{\mathrm{2}} }\left({x}−{B}\right)^{\mathrm{2}} =\left(\frac{{x}}{{B}}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:{y}=\left[\frac{\left(\mathrm{3}+\sqrt{\mathrm{6}\sqrt{\mathrm{2}}}\right){x}}{\mathrm{2}}−\mathrm{1}\right]^{\mathrm{2}} \\ $$
Commented by mr W last updated on 18/Apr/20
Commented by ajfour last updated on 19/Apr/20
$${Thanks},\:{Sir}. \\ $$