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Question Number 53828 by ajfour last updated on 26/Jan/19

Commented by ajfour last updated on 26/Jan/19

$F i n d a a n d b o f m a x i m u m a r e a e l l i p s e$ $w i t h i n t h e t r a p e z i u m .$
	Answered by ajfour last updated on 26/Jan/19

	Commented by ajfour last updated on 26/Jan/19

	$L e t B (h, k), \tan α = 2$ $C (h + 5 s \cos θ, k + 5 s \sin θ)$ $A (h - 4 s \sin θ, k + 4 s \cos θ)$ $D (h - 4 s \sin θ + 3 s \cos θ, k + 4 s \cos θ + 3 s \sin θ)$ $l e t \tan θ = m$ $y_{i n t, B C} = k - m h$ $y_{i n t, A B} = k + \frac{h}{m}$ $y_{i n t, A D} = k + 4 s \cos θ - m (h - 4 s \sin θ)$ $y_{i n t, C D} = k + 5 s \sin θ - (\frac{m - 2}{1 + 2 m}) (h + 5 s \cos θ)$ $d u e t o t a n g e n c y o f A B, B C, C D, A D$ $k - m h = - \sqrt{a^{2} m^{2} + b^{2}} . . . . (i)$ $k + \frac{m}{h} = - \sqrt{\frac{a^{2}}{m^{2}} + b^{2}} . . . . (i i)$ $k + 4 s \cos θ - m (h - 4 s \sin θ) = \sqrt{a^{2} m^{2} + b^{2}}$ $. . . . (i i i)$ $k + 5 s \sin θ - (\frac{m - 2}{1 + 2 m}) (h + 5 s \cos θ)$ $= \sqrt{a^{2} {(\frac{m - 2}{1 + 2 m})}^{2} + b^{2}} . . . (i v)$ $A n d f o r m a x i m u m a r e a o f e l l i p s e$ $d (a b) = 0 . . . (v)$ $U n k n o w n s a r e h, k, m, a, b .$
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