Question

If x=4t/1+t^2 and y=3{1-t^2/1+t^2} then show that dy/dx= -9x/4y

Answer 1

Solution:-

:- We have

$\rm \implies \: x = \dfrac{4t}{1 + {t}^{2} } \: \: \: and \: \: y =3 \bigg( \dfrac{1 - {t}^{2} }{1 + {t}^{2} } \bigg)$

Now Let

$\rm \implies \: t = \tan \theta$

We get

$\rm \implies x = \dfrac{4 \tan\theta }{1 + { \tan}^{2} \theta } \: and \: y = 3 \bigg( \dfrac{1 - \tan^{2} \theta}{ 1+ \tan ^{2} \theta } \bigg)$

we know that Trigonometric identities

$\rm \to \: \dfrac{2 \tan \theta}{1 + \tan ^{2} \theta} = \sin2 \theta \: and \: \: \: \dfrac{1 - \tan^{2} \theta }{1 + \tan^{2} \theta } = \cos2 \theta$

Apply this identity

$\rm \implies \: x = 2 \sin \theta \: \: and \: y = 3 \cos 2 \theta$

Now different x and y with respect to theta

$\rm \implies \: \dfrac{dx}{d \theta} = 4 \cos 2 \theta \: \: and \: \: \dfrac{dy}{d \theta} = - 6 \sin2 \theta$

Now we have to find dy/dx

We can write as

$\rm \implies \dfrac{ \dfrac{dy}{d \theta} }{ \dfrac{dx}{d \theta} } = \dfrac{dy}{d x} = \dfrac{ - 6 \sin2 \theta }{4 \cos2\theta}$

Now take

$\rm \implies \: x = 2 \sin 2\theta$

Multiply by 3 on both side

$\rm \implies \: 3x = 6 \sin2 \theta$

Now take

$\rm \implies \: y = 3 \cos 2 \theta$

Multiply on both side by 4

$\rm \implies4y = 4 \times 3 \cos 2 \theta$

$\rm \implies \: \dfrac{4}{3} y = 4 \cos2 \theta$

Now put the value on

$\rm \implies \dfrac{dy}{d x} = \dfrac{ - 6 \sin2 \theta }{4 \cos2\theta}$

$\rm \implies \dfrac{dy}{dx} = \dfrac{ - 3x}{ \dfrac{4y}{3} } = \dfrac{ - 9x}{4y}$

Hence proved