If x^2 + y^2 = t - 1 t , x^4 + y^4 = t^2 + 1 t^2 then dy dx equal… - Vidyadip

MCQ

If ${x^2} + {y^2} = t - \frac{1}{t},{x^4} + {y^4} = {t^2} + \frac{1}{{{t^2}}}$ then $\frac{{dy}}{{dx}}$ equals

A
$1/x{y^3}$
✓
$1/{x^3}y$
C
$ - 1/{x^3}y$
D
$ - 1/x{y^3}$

✓

Answer

Correct option: B.

$1/{x^3}y$

b
(b) ${x^2} + {y^2} = t - \frac{1}{t}$ and ${x^4} + {y^4} = {t^2} + \frac{1}{{{t^2}}}$
${({x^2})^2} + {({y^2})^2} = {t^2} + \frac{1}{{{t^2}}}$
==> ${({x^2} + {y^2})^2} - 2{x^2}{y^2} = {t^2} + \frac{1}{{{t^2}}}$
==> ${\left( {t - \frac{1}{t}} \right)^2} - 2{x^2}{y^2} = {t^2} + \frac{1}{{{t^2}}}$ ==> ${x^2}{y^2} = - 1$ ==>${y^2} = \frac{{ - 1}}{{{x^2}}}$
Differentiation w.r.t. $x$, we get $2y\frac{{dy}}{{dx}} = \frac{2}{{{x^3}}} \Rightarrow \frac{{dy}}{{dx}} = \frac{1}{{{x^3}y}}$.

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