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Question Number 131642 by liberty last updated on 07/Feb/21
What is the maximum area of ellipse (x^2 /a^2 )+(y^2 /b^2 )=1 which touches the line y = 3x+2.
$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{area}\: \\ $$$$\mathrm{of}\:\mathrm{ellipse}\:\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} }+\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{b}^{\mathrm{2}} }=\mathrm{1}\:\mathrm{which}\:\mathrm{touches} \\ $$$$\mathrm{the}\:\mathrm{line}\:\mathrm{y}\:=\:\mathrm{3x}+\mathrm{2}. \\ $$
Answered by benjo_mathlover last updated on 07/Feb/21
Area ellipse = πab touches the line y=3x+2 by tangency ⇒y=mx+(√(a^2 m^2 +b^2 )) where { ((m=3)),(((√(9a^2 +b^2 )) = 2)) :} or b=(√(4−9a^2 )) then A=f(a)=πa(√(4−9a^2 )) A^2 = π^2 a^2 (4−9a^2 ) 9π^2 (a^2 )^2 −4π^2 a^2 +A^2 =0 Δ = 16π^4 −4(9π^2 )A^2 =0 ⇔ 4π^2 =9A^2 ⇒A=((2π)/3) ←max area
$$\mathrm{Area}\:\mathrm{ellipse}\:=\:\pi\mathrm{ab} \\ $$$$\mathrm{touches}\:\mathrm{the}\:\mathrm{line}\:\mathrm{y}=\mathrm{3x}+\mathrm{2} \\ $$$$\mathrm{by}\:\mathrm{tangency}\:\Rightarrow\mathrm{y}=\mathrm{mx}+\sqrt{\mathrm{a}^{\mathrm{2}} \mathrm{m}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} } \\ $$$$\mathrm{where}\:\begin{cases}{\mathrm{m}=\mathrm{3}}\\{\sqrt{\mathrm{9a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} }\:=\:\mathrm{2}}\end{cases} \\ $$$$\mathrm{or}\:\mathrm{b}=\sqrt{\mathrm{4}−\mathrm{9a}^{\mathrm{2}} } \\ $$$$\mathrm{then}\:\mathrm{A}=\mathrm{f}\left(\mathrm{a}\right)=\pi\mathrm{a}\sqrt{\mathrm{4}−\mathrm{9a}^{\mathrm{2}} } \\ $$$$\mathrm{A}^{\mathrm{2}} \:=\:\pi^{\mathrm{2}} \mathrm{a}^{\mathrm{2}} \left(\mathrm{4}−\mathrm{9a}^{\mathrm{2}} \right) \\ $$$$\mathrm{9}\pi^{\mathrm{2}} \left(\mathrm{a}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{4}\pi^{\mathrm{2}} \mathrm{a}^{\mathrm{2}} +\mathrm{A}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Delta\:=\:\mathrm{16}\pi^{\mathrm{4}} −\mathrm{4}\left(\mathrm{9}\pi^{\mathrm{2}} \right)\mathrm{A}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Leftrightarrow\:\mathrm{4}\pi^{\mathrm{2}} =\mathrm{9A}^{\mathrm{2}} \:\Rightarrow\mathrm{A}=\frac{\mathrm{2}\pi}{\mathrm{3}}\:\leftarrow\mathrm{max}\:\mathrm{area} \\ $$
Answered by mr W last updated on 07/Feb/21
area of ellipse is A=πab since it touches the line 3x−y+2=0, (3a)^2 +(−1×b)^2 =2^2 ⇒b^2 =4−9a^2 A=π(√(a^2 b^2 ))=π(√(a^2 (4−9a^2 )))=(π/3)(√(9a^2 (4−9a^2 ))) ≤(π/3)×((9a^2 +(4−9a^2 ))/2)=((2π)/3) i.e. A_(max) =((2π)/3) when 9a^2 =4−9a^2 ⇒a=((√2)/3) ⇒b=3a=(√2)
$${area}\:{of}\:{ellipse}\:{is} \\ $$$${A}=\pi{ab} \\ $$$${since}\:{it}\:{touches}\:{the}\:{line}\:\mathrm{3}{x}−{y}+\mathrm{2}=\mathrm{0}, \\ $$$$\left(\mathrm{3}{a}\right)^{\mathrm{2}} +\left(−\mathrm{1}×{b}\right)^{\mathrm{2}} =\mathrm{2}^{\mathrm{2}} \\ $$$$\Rightarrow{b}^{\mathrm{2}} =\mathrm{4}−\mathrm{9}{a}^{\mathrm{2}} \\ $$$$ \\ $$$${A}=\pi\sqrt{{a}^{\mathrm{2}} {b}^{\mathrm{2}} }=\pi\sqrt{{a}^{\mathrm{2}} \left(\mathrm{4}−\mathrm{9}{a}^{\mathrm{2}} \right)}=\frac{\pi}{\mathrm{3}}\sqrt{\mathrm{9}{a}^{\mathrm{2}} \left(\mathrm{4}−\mathrm{9}{a}^{\mathrm{2}} \right)} \\ $$$$\leqslant\frac{\pi}{\mathrm{3}}×\frac{\mathrm{9}{a}^{\mathrm{2}} +\left(\mathrm{4}−\mathrm{9}{a}^{\mathrm{2}} \right)}{\mathrm{2}}=\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$$${i}.{e}.\:{A}_{{max}} =\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$$${when}\:\mathrm{9}{a}^{\mathrm{2}} =\mathrm{4}−\mathrm{9}{a}^{\mathrm{2}} \:\Rightarrow{a}=\frac{\sqrt{\mathrm{2}}}{\mathrm{3}}\:\Rightarrow{b}=\mathrm{3}{a}=\sqrt{\mathrm{2}} \\ $$
Answered by ajfour last updated on 07/Feb/21
let x=pt ⇒ y=3pt+2 let 3p=1 ⇒ (t^2 /(9a^2 ))+(y^2 /b^2 )=1 Now max area when r=3a=b (√2)=r=b=3a A_(max) =((2π)/3)
$${let}\:\:{x}={pt} \\ $$$$\Rightarrow\:\:{y}=\mathrm{3}{pt}+\mathrm{2} \\ $$$${let}\:\:\mathrm{3}{p}=\mathrm{1} \\ $$$$\Rightarrow\:\:\frac{{t}^{\mathrm{2}} }{\mathrm{9}{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$${Now}\:{max}\:{area}\:\:{when}\:\:{r}=\mathrm{3}{a}={b} \\ $$$$\sqrt{\mathrm{2}}={r}={b}=\mathrm{3}{a} \\ $$$${A}_{{max}} =\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$
Commented by ajfour last updated on 07/Feb/21
Commented by ajfour last updated on 07/Feb/21
3x=t
$$\mathrm{3}{x}={t} \\ $$

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