Question #1b68f

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Answer 1 · 2017-04-06T14:49:28

$dy/dx=-x/sqrt(tan(a^2)-x^2)$

Applying $tan$ to both sides

$x^2+y^2=tan(a^2)$

now deriving

$2xdx+2ydy=0$ or

$dy/dx = -x/y$

but

$y = sqrt(tan(a^2)-x^2)$

so finally

$dy/dx=-x/sqrt(tan(a^2)-x^2)$