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if the area bounded by x-axis and $y=mx$ and ellipse $\displaystyle \frac{x^2}{16}+\frac{y^2}{9} = 1$ in first quadrant is

$\displaystyle \frac{1}{8}^{th}$ of the area of ellipse. then $m$ is

Attempt: solving $y=mx$ and $\displaystyle \frac{x^2}{16}+\frac{y^2}{9} = 1$ we have $9x^2+16m^2x^2=144$

$\displaystyle x^2=\frac{144}{9+16m^2}$ so $\displaystyle x = \frac{12}{\sqrt{9+16m^2}}=k$

so $\displaystyle \int^{k}_{0}mxdx+\frac{3}{4}\int^{4}_{k}\sqrt{16-x^2}dx = \frac{12 \pi}{8} $

$\displaystyle \frac{mk^2}{2}+\frac{3}{4}\bigg[\frac{x}{2}\sqrt{16-x^2}+8\sin^{-1}\left(\frac{x}{4}\right)\bigg]\bigg|_{k}^{4} = \frac{3\pi}{2}$

$$\frac{mk^2}{2}+\frac{3}{4}\bigg[4\pi-\frac{k}{2}\sqrt{16-k^2}+\frac{8}{k}\sin^{-1}\left(\frac{k}{4}\right)\bigg] = \frac{3\pi}{2}$$

could some help me to solve this, thanks

1 Answers 1

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$\frac {x^2}{a^2} + \frac {y^2}{b^2} = 1$

Tranform the coordinate system as follows. $x = a u, y = b v$

$u^2 + v^2 = 1$ Our ellipse has become a circle.

$y = mx$ becomes $v = \frac {am}{b} u$

The boundaries of the u axis and the line $v = \frac {am}{b} u$ define $\frac 18$ circle

$\frac {am}{b} = 1\\ m = \frac ba = \frac 34$

Cleaning up what you have already done:

$9k^2 + 16m^2k^2 = 144\\ 16m^2k^2 = 9(16^2-k^2)\\ mk = \frac {3}{4}\sqrt{16^2 - k^2}$

$\frac 12mk^2 - \frac{3k}{8}\sqrt{16-k^2} = 0$

$3\pi - 6\arcsin\frac k4 = \frac {3\pi}{2}$