Question Number 30377 by ajfour last updated on 21/Feb/18
Commented by ajfour last updated on 21/Feb/18
$${Given}\:{the}\:{ellipse}\:{touches}\:{parabola} \\ $$$${only}\:{at}\:{vertex}\:{and}\:{lies}\:{within}\:{the} \\ $$$${the}\:{red}\:{line}\:{and}\:{parabola}. \\ $$
Answered by mrW2 last updated on 22/Feb/18
![Eqn. of ellipse: ((x/a))^2 +(((2y)/x_0 )−1)^2 =1 Intersection with parabola: ((x_0 y)/a^2 )=1−(((2y)/x_0 )−1)^2 =((4y)/x_0 )(1−(y/x_0 )) y[(x_0 /a^2 )−(4/x_0 )+((4y)/x_0 ^2 )]=0 ⇒y=0 ⇒(x_0 /a^2 )−(4/x_0 )+((4y)/x_0 ^2 )=0 ⇒y=x_0 −(x_0 ^3 /(4a^2 ))=x_0 (1−(x_0 ^2 /(4a^2 )))=^! 0 ⇒(2a)^2 =x_0 ^2 ⇒a=(x_0 /2) i.e. the largest ellipse is a circle: (((2x)/x_0 ))^2 +(((2y)/x_0 )−1)^2 =1 with radius (x_0 /2) and its area is ((πx_0 ^2 )/4).](Q30383.png)
$${Eqn}.\:{of}\:{ellipse}: \\ $$$$\left(\frac{{x}}{{a}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{2}{y}}{{x}_{\mathrm{0}} }−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$$ \\ $$$${Intersection}\:{with}\:{parabola}: \\ $$$$\frac{{x}_{\mathrm{0}} {y}}{{a}^{\mathrm{2}} }=\mathrm{1}−\left(\frac{\mathrm{2}{y}}{{x}_{\mathrm{0}} }−\mathrm{1}\right)^{\mathrm{2}} =\frac{\mathrm{4}{y}}{{x}_{\mathrm{0}} }\left(\mathrm{1}−\frac{{y}}{{x}_{\mathrm{0}} }\right) \\ $$$${y}\left[\frac{{x}_{\mathrm{0}} }{{a}^{\mathrm{2}} }−\frac{\mathrm{4}}{{x}_{\mathrm{0}} }+\frac{\mathrm{4}{y}}{{x}_{\mathrm{0}} ^{\mathrm{2}} }\right]=\mathrm{0} \\ $$$$\Rightarrow{y}=\mathrm{0} \\ $$$$\Rightarrow\frac{{x}_{\mathrm{0}} }{{a}^{\mathrm{2}} }−\frac{\mathrm{4}}{{x}_{\mathrm{0}} }+\frac{\mathrm{4}{y}}{{x}_{\mathrm{0}} ^{\mathrm{2}} }=\mathrm{0} \\ $$$$\Rightarrow{y}={x}_{\mathrm{0}} −\frac{{x}_{\mathrm{0}} ^{\mathrm{3}} }{\mathrm{4}{a}^{\mathrm{2}} }={x}_{\mathrm{0}} \left(\mathrm{1}−\frac{{x}_{\mathrm{0}} ^{\mathrm{2}} }{\mathrm{4}{a}^{\mathrm{2}} }\right)\overset{!} {=}\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{2}{a}\right)^{\mathrm{2}} ={x}_{\mathrm{0}} ^{\mathrm{2}} \\ $$$$\Rightarrow{a}=\frac{{x}_{\mathrm{0}} }{\mathrm{2}} \\ $$$${i}.{e}.\:{the}\:{largest}\:{ellipse}\:{is}\:{a}\:{circle}: \\ $$$$\left(\frac{\mathrm{2}{x}}{{x}_{\mathrm{0}} }\right)^{\mathrm{2}} +\left(\frac{\mathrm{2}{y}}{{x}_{\mathrm{0}} }−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$${with}\:{radius}\:\frac{{x}_{\mathrm{0}} }{\mathrm{2}}\:{and}\:{its}\:{area}\:{is}\:\frac{\pi{x}_{\mathrm{0}} ^{\mathrm{2}} }{\mathrm{4}}. \\ $$
Commented by ajfour last updated on 21/Feb/18
$${Good}\:{way}\:{Sir}.\: \\ $$