Question

if x2 + y2 = t + 1/t and x4 +y4 = t2 + 1/t2, then prove tht dy/dx = 1/ x3y

Answer 1

For Calculation see Pic:


Hope, you understand my answer!!

Answer 2

$if \: \: {x}^{2} + {y}^{2} = t + \frac{1}{t} \\ \\ and \: \: {x}^{4} + {y}^{4} = {t}^{2} + \frac{1}{ {t}^{2} }$
squaring both sides,the equation given below


${x}^{2} + {y}^{2} = t + \frac{1}{t}$
we get

$( {x^{2} + {y}^{2} })^{2} = {(t - \frac{1}{t} )}^{2} \\ \\ {x}^{4} + {y}^{4} + 2 {x}^{2} {y}^{2} = {t}^{2} + \frac{1}{ {t}^{2} } - 2t( \frac{1}{t} )$
from another equation put the value of
${x}^{4} + {y}^{4} \: \: in \: the \: eq$
${t}^{2} + \frac{1}{ {t}^{2} } + 2 {x}^{2} {y}^{2} = {t}^{2} + \frac{1}{ {t}^{2} } - 2\\ \\ \\ 2 {x}^{2} {y}^{2} = - 2 \\ \\ {x}^{2} {y}^{2} = - 1....eq1 \\ \\ or \\ \\ y = \frac{ - 1}{ {x}^{2}y } ......eq2$
take derivative of eq1 with respect to x

$2x {y}^{2} + 2 {x}^{2} y \: \frac{dy}{dx} = 0 \\ \\ 2 {x}^{2} y \: \frac{dy}{dx} = 2x {y}^{2} \\ \\ \frac{dy}{dx} = \frac{y}{x} \\ \\ put \: value \: of \: y \: from \: eq2 \\ \\ \frac{dy}{dx} = ( \frac{1}{x} )(\frac{ - 1}{ {x}^{2}y }) \\ \\ \frac{dy}{dx} = \frac{ - 1}{ {x}^{3}y }$
Hence proved