Question

Please solve this question​

Answer 1

$x = \frac{2at}{1 + {t}^{2} } \: \: \: \: \: \: \: \: \: \: y = \frac{1 - {t}^{2} }{1 + {t}^{2} } \\ put \: \: \: t = tan \theta \\ \\ x = \frac{a2tan \theta}{1 + {tan}^{2} \theta } \\ x = asin2 \theta \\ \frac{dx}{d \theta} = 2acos2 \theta \\ \\ y = \frac{1 - {tan}^{2} \theta }{1 + {tan}^{2} \theta } \\ y = cos2 \theta \\ \frac{dy}{d \theta} = - 2sin2 \theta \\ \\ \therefore \: \: \: \: \frac{dy}{dx} = \frac{dy}{d \theta} \times \frac{d \theta}{dx} \\ \\ = \frac{ - 2sin 2\theta}{2acos2 \theta} \\ \\ = - \frac{1}{a} tan2 \theta \\ \\ = - \frac{1}{a} \frac{2tan \theta}{1 - {tan}^{2} \theta} \\ \\ = - \frac{1}{a} \frac{2t}{1 - {t}^{2} }$