Question

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Answer 1

Ответ:

по формуле:

$y'x = \frac{y't}{x't}$

$y't = \frac{1}{ { \cos}^{2} 2t} \times 2$

$x't = - 2a \sin(2t)$

$y'x = \frac{2}{ { \cos}^{2}2t } \times \frac{1}{( - 2a \sin(2t)) } = \\ = - \frac{1}{a \sin(2t) { \cos }^{2} (2t) }$