# Question-66055

Question Number 66055 by ajfour last updated on 08/Aug/19
Commented by ajfour last updated on 08/Aug/19
$${Eq}.\:{of}\:{ellipse}\:\:\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{\left({y}−{b}\right)^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\$$$${parabola}\:{touches}\:{the}\:{x}-{axis} \\$$$${and}\:{the}\:{ellipse}\:{and}\:{passes} \\$$$${though}\:{center}\:{of}\:{ellipse}\:\left(\mathrm{0},{b}\right). \\$$$${Find}\:{equation}\:{of}\:{parabola}\:{in} \\$$$${terms}\:{of}\:{a}\:{and}\:{b}. \\$$
Answered by mr W last updated on 09/Aug/19
$${eqn}.\:{of}\:{parabola}: \\$$$${y}={b}\left(\mathrm{1}−\frac{{x}}{{p}}\right)^{\mathrm{2}} \\$$$${eqn}.\:{of}\:{ellipse}: \\$$$$\left(\frac{{x}}{{a}}\right)^{\mathrm{2}} +\left(\frac{{y}}{{b}}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{1} \\$$$${touch}\:{point}\:{P}\left({h},{k}\right) \\$$$${let}\:{m}=\mathrm{tan}\:\theta\:{for}\:{tangent}\:{at}\:{touch}\:{point} \\$$$${y}'=−\frac{\mathrm{2}{b}}{{p}}\left(\mathrm{1}−\frac{{h}}{{p}}\right)={m} \\$$$$\Rightarrow{h}={p}\left(\mathrm{1}+\frac{{pm}}{\mathrm{2}{b}}\right) \\$$$$\frac{{x}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}}\left(\frac{{y}}{{b}}−\mathrm{1}\right){y}'=\mathrm{0} \\$$$$\frac{{h}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}}\left(\frac{{k}}{{b}}−\mathrm{1}\right){m}=\mathrm{0} \\$$$$\Rightarrow{k}={b}\left[\mathrm{1}−\frac{{pb}}{{ma}^{\mathrm{2}} }\left(\mathrm{1}+\frac{{pm}}{\mathrm{2}{b}}\right)\right] \\$$$$\left(\frac{{h}}{{a}}\right)^{\mathrm{2}} +\left(\frac{{k}}{{b}}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{1} \\$$$$\Rightarrow\frac{{p}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\left(\mathrm{1}+\frac{{b}^{\mathrm{2}} }{{m}^{\mathrm{2}} {a}^{\mathrm{2}} }\right)\left(\mathrm{1}+\frac{{pm}}{\mathrm{2}{b}}\right)^{\mathrm{2}} =\mathrm{1}\:\:\:…\left({i}\right) \\$$$$\\$$$$\frac{{k}}{{b}}=\left(\mathrm{1}−\frac{{h}}{{p}}\right)^{\mathrm{2}} \\$$$$\Rightarrow\mathrm{1}−\frac{{pb}}{{ma}^{\mathrm{2}} }\left(\mathrm{1}+\frac{{pm}}{\mathrm{2}{b}}\right)=\left(\frac{{pm}}{\mathrm{2}{b}}\right)^{\mathrm{2}} \:\:\:…\left({ii}\right) \\$$$${let}\:\lambda=\frac{{p}}{{a}},\:\mu=\frac{{b}}{{a}} \\$$$${from}\:\left({i}\right): \\$$$$\Rightarrow\lambda^{\mathrm{2}} \left(\mathrm{1}+\frac{\mu^{\mathrm{2}} }{{m}^{\mathrm{2}} }\right)\left(\mathrm{1}+\frac{{m}\lambda}{\mathrm{2}\mu}\right)^{\mathrm{2}} =\mathrm{1}\:\:\:\:…\left({I}\right) \\$$$${from}\:\left({ii}\right): \\$$$$\Rightarrow\mathrm{1}−\frac{\mu\lambda}{{m}}\left(\mathrm{1}+\frac{{m}\lambda}{\mathrm{2}\mu}\right)=\left(\frac{{m}\lambda}{\mathrm{2}\mu}\right)^{\mathrm{2}} \\$$$$\Rightarrow\left(\mathrm{2}+\frac{{m}^{\mathrm{2}} }{\mu^{\mathrm{2}} }\right)\lambda^{\mathrm{2}} +\frac{\mathrm{4}\mu}{{m}}\lambda−\mathrm{4}\:=\mathrm{0}\:\:\:…\left({II}\right) \\$$$$\Rightarrow\lambda=\frac{\frac{\mathrm{2}{m}}{\mu}}{\mathrm{2}+\frac{{m}^{\mathrm{2}} }{\mu^{\mathrm{2}} }}=\frac{\mathrm{2}\mu{m}}{\mathrm{2}\mu^{\mathrm{2}} +{m}^{\mathrm{2}} } \\$$$${or}\:{m}^{\mathrm{2}} −\frac{\mathrm{2}\mu}{\lambda}{m}+\mathrm{2}\mu^{\mathrm{2}} =\mathrm{0} \\$$$$\Rightarrow\frac{{m}}{\mu}=\frac{\mathrm{1}}{\lambda}\left(\mathrm{1}+\sqrt{\mathrm{1}−\mathrm{2}\lambda^{\mathrm{2}} }\right) \\$$$${put}\:{this}\:{into}\:\left({I}\right): \\$$$$\Rightarrow\frac{\lambda^{\mathrm{2}} \left(\mathrm{3}+\sqrt{\mathrm{1}−\mathrm{2}\lambda^{\mathrm{2}} }\right)^{\mathrm{2}} }{\mathrm{8}}\left(\mathrm{2}+\frac{\lambda^{\mathrm{2}} }{\mathrm{1}−\lambda^{\mathrm{2}} +\sqrt{\mathrm{1}−\mathrm{2}\lambda^{\mathrm{2}} }}\right)=\mathrm{1} \\$$$$\Rightarrow\lambda=\mathrm{0}.\mathrm{5194}\:\left({constant},\:{independent}\:{from}\:\mu!\right) \\$$$$\Rightarrow{p}=\lambda{a}=\mathrm{0}.\mathrm{5194}{a} \\$$$$\\$$$${eqn}.\:{of}\:{parabola}: \\$$$$\Rightarrow{y}={b}\left(\mathrm{1}−\frac{{x}}{\mathrm{0}.\mathrm{5194}{a}}\right)^{\mathrm{2}} \\$$
Commented by MJS last updated on 09/Aug/19
$$\mathrm{great}! \\$$$$\mathrm{just}\:\mathrm{for}\:\mathrm{the}\:\mathrm{fun}\:\mathrm{of}\:\mathrm{it},\:\mathrm{the}\:\mathrm{exact}\:\mathrm{value}\:\mathrm{of}\:\lambda: \\$$$$\lambda=\sqrt{−\mathrm{4}+\sqrt[{\mathrm{3}}]{\mathrm{56}+\frac{\mathrm{88}}{\mathrm{9}}\sqrt{\mathrm{33}}}−\sqrt[{\mathrm{3}}]{−\mathrm{56}+\frac{\mathrm{88}}{\mathrm{9}}\sqrt{\mathrm{33}}}} \\$$
Commented by mr W last updated on 08/Aug/19
Commented by mr W last updated on 08/Aug/19
Commented by mr W last updated on 09/Aug/19
$${thanks}\:{sir}!\:{how}\:{did}\:{you}\:{get}\:{that}? \\$$$${could}\:{you}\:{transform}\:{the}\:{eqn}.\:{into}\:{a} \\$$$${cubic}\:{eqn}.? \\$$
Commented by MJS last updated on 12/Aug/19
$$\mathrm{sorry}\:\mathrm{I}\:\mathrm{get}\:\mathrm{no}\:\mathrm{notifications}\:\mathrm{again}… \\$$$$\mathrm{it}'\mathrm{s}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{transform}\:\mathrm{to} \\$$$$\lambda^{\mathrm{6}} +\mathcal{A}\lambda^{\mathrm{4}} +\mathcal{B}\lambda^{\mathrm{2}} +\mathcal{C}=\mathrm{0} \\$$$$\mathrm{and}\:\mathrm{then}\:\mathrm{follow}\:\mathrm{the}\:\mathrm{usual}\:\mathrm{path} \\$$
Commented by mr W last updated on 14/Aug/19
$${now}\:{i}\:{see}\:{sir}!\:{thanks}! \\$$$${i}\:{get}\:{no}\:{notification}\:{either}.\::\left(\right. \\$$