Question

If x=4t/1+t^2, y=3(1-t^2) /(1+t^2), then show that dy/dx=-9x/4y​

Answer 1

SOLUTION

GIVEN

$\displaystyle \sf{x = \frac{4t}{1 + {t}^{2} } \: \: \: and \: \: \: y = \frac{3(1 - {t}^{2} )}{1 + {t}^{2} } }$

TO PROVE

$\displaystyle \sf{ \frac{dy}{dx} = - \frac{9x}{4y} }$

EVALUATION

Here it is given that

$\displaystyle \sf{x = \frac{4t}{1 + {t}^{2} } \: \: \: and \: \: \: y = \frac{3(1 - {t}^{2} )}{1 + {t}^{2} } }$

Now Squaring both we get

$\displaystyle \sf{ {x}^{2} = \frac{16 {t}^{2} }{{(1 + {t}^{2} )}^{2} } \: }$

$\displaystyle \sf{ {y}^{2} = \frac{9{(1 - {t}^{2} )}^{2} }{{(1 + {t}^{2})}^{2} } }$

Now

$\sf{9 {x}^{2} + 4 {y}^{2} }$

$\displaystyle \sf{ = \frac{144 {t}^{2} }{{(1 + {t}^{2} )}^{2} } + \frac{36{(1 - {t}^{2} )}^{2} }{{(1 + {t}^{2})}^{2} } \: }$

$\displaystyle \sf{ = \frac{36 \bigg[{(1 - {t}^{2} )}^{2} + 4 {t}^{2} \bigg] }{{(1 + {t}^{2})}^{2} } \: }$

$\displaystyle \sf{ = \frac{36 \bigg[{(1 + {t}^{2} )}^{2} \bigg] }{{(1 + {t}^{2})}^{2} } \: }$

$= 36$

$\sf{ \therefore \: \: 9 {x}^{2} + 4 {y}^{2} = 36 }$

Differentiating both sides with respect to x we get

$\displaystyle \sf{18x + 8y \frac{dy}{dx} = 0 }$

$\displaystyle \sf{ \implies \: \frac{dy}{dx} = - \frac{18x}{8y} }$

$\displaystyle \sf{ \implies \: \frac{dy}{dx} = - \frac{9x}{4y} }$

Hence proved

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