If x²+y²={t}+{1}{{t}}, x^{4}+y^{4}={t}²+{1}{{t}²}, then d y/d x=

Correct Answer is : −1x3y-{1}{x³ y}−x3y1

Correct Answer is : −1x3y-{1}{x³ y}−x3y1

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MHTCET 11 May 2024 Shift 1 Paper

Question : 121 of 150

Marks: +1, -0

x^{2}+y^{2}=\mathrm{t}+\frac{1}{\mathrm{t}}, x^{4}+y^{4}=\mathrm{t}^{2}+\frac{1}{\mathrm{t}^{2}}

\frac{d y}{d x}=

Solution:

Concept:

Use the given equations to eliminate

t

and obtain a relation between

x

and

y

, then differentiate implicitly.

Explanation:

Given

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

and

x^{4}+y^{4}=t^{2}+\frac{1}{t^{2}}

Square the first equation:

(x^{2}+y^{2})^{2}=\left(t+\frac{1}{t}\right)^{2}

x^{4}+y^{4}+2x^{2}y^{2}=t^{2}+\frac{1}{t^{2}}+2

Substitute

x^{4}+y^{4}=t^{2}+\frac{1}{t^{2}}

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

2x^{2}y^{2}=2

x^{2}y^{2}=1

y^{2}=\frac{1}{x^{2}}

Differentiate

y^{2}=\frac{1}{x^{2}}

with respect to

x

2y\frac{dy}{dx}=-\frac{2}{x^{3}}

\frac{dy}{dx}=-\frac{1}{x^{3}y}

Answer:

\frac{dy}{dx}=-\frac{1}{x^{3}y}

Hence, the correct option is D.

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