find-the-area-bounded-the-curves-y-2-36-12x-and-y-2-16-8x-

Question Number 102444 by bemath last updated on 09/Jul/20

f i n d t h e a r e a b o u n d e d t h e

c u r v e s y^{2} = 36 + 12 x a n d

y^{2} = 16 - 8 x

Answered by bemath last updated on 09/Jul/20

\Leftrightarrow \frac{y^{2} - 36}{12} = \frac{16 - y^{2}}{8}

\frac{y^{2}}{12} - 3 = 2 - \frac{y^{2}}{8}

\frac{5 y^{2}}{24} - 5 = 0 \Rightarrow Δ = \frac{4.5.5}{24} = \frac{25}{6}

A r e a = \frac{Δ \sqrt{Δ}}{6 . a^{2}} = \frac{\frac{25}{6} . \frac{5}{\sqrt{6}}}{6 . (\frac{25}{24^{2}})}

= \frac{125}{6 \sqrt{6}} \times \frac{24.24}{6.25} = \frac{80}{\sqrt{6}}

= \frac{40 \sqrt{6}}{3} .

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