What-is-the-maximum-area-of-ellipse-x-2-a-2-y-2-b-2-1-which-touches-the-line-y-3x-2-

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 = 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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