What is the maximum area of ellipse (x^2 /a^2 )+(y^2 /b^2 )=1 which touches the line y = 3x+2.


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Question Number 131642 by liberty last updated on 07/Feb/21

$What is the maximum area$ $of ellipse \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 which touches$ $the line y = 3 x + 2 .$
	Answered by benjo_mathlover last updated on 07/Feb/21
	$Area ellipse = πab touches the line y=3x+2 by tangency ⇒y=mx+(√(a^2 m^2 +b^2 )) where { ((m=3)),(((√(9a^2 +b^2 )) = 2)) :} or b=(√(4−9a^2 )) then A=f(a)=πa(√(4−9a^2 )) A^2 = π^2 a^2 (4−9a^2 ) 9π^2 (a^2 )^2 −4π^2 a^2 +A^2 =0 Δ = 16π^4 −4(9π^2 )A^2 =0 ⇔ 4π^2 =9A^2 ⇒A=((2π)/3) ←max area$
	$Area ellipse = π ab$ $touches the line y = 3 x + 2$ $by tangency \Rightarrow y = mx + \sqrt{a^{2} m^{2} + b^{2}}$ $where {\begin{cases} m = 3 \\ \sqrt{{9 a}^{2} + b^{2}} = 2 \end{cases}$ $or b = \sqrt{4 - {9 a}^{2}}$ $then A = f (a) = π a \sqrt{4 - {9 a}^{2}}$ $A^{2} = π^{2} a^{2} (4 - {9 a}^{2})$ $9 π^{2} {(a^{2})}^{2} - 4 π^{2} a^{2} + A^{2} = 0$ $Δ = 16 π^{4} - 4 (9 π^{2}) A^{2} = 0$ $\Leftrightarrow 4 π^{2} = {9 A}^{2} \Rightarrow A = \frac{2 π}{3} \leftarrow \max area$
	Answered by mr W last updated on 07/Feb/21

	$a r e a o f e l l i p s e i s$ $A = π a b$ $s i n c e i t t o u c h e s t h e l i n e 3 x - y + 2 = 0,$ ${(3 a)}^{2} + {(- 1 \times b)}^{2} = 2^{2}$ $\Rightarrow b^{2} = 4 - 9 a^{2}$ $A = π \sqrt{a^{2} b^{2}} = π \sqrt{a^{2} (4 - 9 a^{2})} = \frac{π}{3} \sqrt{9 a^{2} (4 - 9 a^{2})}$ $⩽ \frac{π}{3} \times \frac{9 a^{2} + (4 - 9 a^{2})}{2} = \frac{2 π}{3}$ $i . e . A_{m a x} = \frac{2 π}{3}$ $w h e n 9 a^{2} = 4 - 9 a^{2} \Rightarrow a = \frac{\sqrt{2}}{3} \Rightarrow b = 3 a = \sqrt{2}$
	Answered by ajfour last updated on 07/Feb/21

	$l e t x = p t$ $\Rightarrow y = 3 p t + 2$ $l e t 3 p = 1$ $\Rightarrow \frac{t^{2}}{9 a^{2}} + \frac{y^{2}}{b^{2}} = 1$ $N o w m a x a r e a w h e n r = 3 a = b$ $\sqrt{2} = r = b = 3 a$ $A_{m a x} = \frac{2 π}{3}$
	Commented by ajfour last updated on 07/Feb/21

	Commented by ajfour last updated on 07/Feb/21

	$3 x = t$
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