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Question Number 49394 by ajfour last updated on 06/Dec/18

Commented by ajfour last updated on 07/Dec/18

$$Findequationofmaximumarea$$$$ellipse.$$

Answered by ajfour last updated on 06/Dec/18

$$y=c-{ax}^2$$$$\frac{x^2}{p^2}+\frac{{(y-q)}^2}{q^2}=1$$$$\frac{x^2}{p^2}+\frac{{(c-{ax}^2-q)}^2}{q^2}=1$$$$p^2{a}^2{x}^4+[q^2-2a(c-q)p^2]x^2+p^2[{(c-q)}^2-q^2]=0$$$$sofortangency$$$${[q^2-2a(c-q)p^2]}^2=4a^2{p}^4[{(c-q)}^2-q^2)$$$$\Rightarrow q^4-4a(c-q)p^2{q}^2+4a^2{p}^4{q}^2=0$$$$or{q}^2-4a(c-q)p^2+4a^2{p}^4=0...\left(i\right)$$$$letA=p^2{q}^2\Rightarrow$$$$q^4-4a(c-q)A+\frac{4a^2{A}^2}{q^2}=0$$$$\Rightarrow 4a^2{A}^2-4{aq}^2(c-q)A+q^6=0$$$$\Rightarrow \left(8a^{2}A\right)\frac{dA}{dq}-4{aq}^2(c-q)\frac{dA}{dq}$$$$+(12{aq}^2-8acq)A+6q^5=0$$$$\frac{dA}{dq}=0\Rightarrow$$$$A(12{aq}^2-8acq)=-6q^5$$$$\Rightarrow 2{ap}^2(2c-3q)=3q^2....\left(ii\right)$$$$using{p}^2=\frac{3q^2}{2a(2c-3q)}in..\left(i\right)$$$$q^2-4a(c-q)\left[\frac{3q^2}{2a(2c-3q)}\right]+4a^2{\left[\frac{3q^2}{2a(2c-3q)}\right]}^2=0$$$$\Rightarrow$$$$4a^2{q}^2{(2c-3q)}^2-24a^2{q}^2(c-q)(2c-3q)$$$$36a^2{q}^4=0$$$$\Rightarrow {(2c-3q)}^2-6(c-q)(2c-3q)$$$$+9q^2=0$$$$\Rightarrow 9{\mathit{q}}^2+4{\mathit{c}}^2-12\mathit{c}\mathit{q}-12{\mathit{c}}^2+30\mathit{c}\mathit{q}$$$$-18{\mathit{q}}^2+9{\mathit{q}}^2=0$$$$\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}$$$$\Rightarrow q=\frac{4c}{9}andfrom\left(i\right),$$$$p^2=\frac{3q^2}{2a(2c-3q)}=\frac{\left(\frac{16c^2}{27}\right)}{2a\left(\frac{2c}{3}\right)}=\frac{4c}{9a}$$$$eq.ofellipseisthen$$$$\frac{9{\mathit{a}\mathit{x}}^2}{4\mathit{c}}+\frac{81{(\mathit{y}-\frac{4\mathit{c}}{9})}^2}{16{\mathit{c}}^2}=1$$$$\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}.$$

Commented by mr W last updated on 06/Dec/18

$$veryfinesir!$$

Commented by ajfour last updated on 06/Dec/18

$$yessir,wouldyoucheckthe$$$$solutionagain,please..$$

Commented by mr W last updated on 06/Dec/18

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