Question Number 49526 by ajfour last updated on 07/Dec/18
Commented by ajfour last updated on 07/Dec/18
$${If}\:{both}\:{the}\:{coloured}\:{areas}\:{are}\:{equal}, \\ $$$${find}\:{equation}\:{of}\:{parabola}\:{in}\:{terms} \\ $$$${of}\:{ellipse}\:{parameters}\:\boldsymbol{{a}}\:{and}\:\boldsymbol{{b}}. \\ $$
Answered by ajfour last updated on 07/Dec/18
$${y}=\:{Ax}^{\mathrm{2}} −{b}\:\:\:\:,\:\:\:\:\:\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\:=\:\mathrm{1} \\ $$$$\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{\left({Ax}^{\mathrm{2}} −{b}\right)^{\mathrm{2}} }{{b}^{\mathrm{2}} }\:=\:\mathrm{1} \\ $$$${A}^{\mathrm{2}} {x}^{\mathrm{4}} +\left(\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }−\mathrm{2}{Ab}\right){x}^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\Rightarrow\:\:{x}_{{P}} \:=\:\frac{\sqrt{\mathrm{2}{Ab}−\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}}{{A}} \\ $$$$\Rightarrow\:\int_{\mathrm{0}} ^{\:\:{x}_{{P}} } \:\left({b}\sqrt{\mathrm{1}−\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}−{Ax}^{\mathrm{2}} +{b}\right){dx}\:=\frac{\pi{ab}}{\mathrm{4}}\:. \\ $$$$ \\ $$