NIMCET 2019 Question Paper with solutions

Question : 36 of 120

Marks: +1, -0

The tangent at the point (2, -2) to the curve

x^2y^2—2x=4(1-y)

does not pass through the point___

Solution:

We have

x^2y^2-2x=4(1-y)

On differentiating with respect to x we get

2x^{2}y\;\;{dy}/{dx}+2xy^{2}-2=-4\;\;{dy}/{dx}

\;\;{dy}/{dx}=\;\;{2-2xy^{2}}/{2x^{2}y+4}=\;\;{1-xy^{2}}/{x^{2}y+2}

(\;\;{dy}/{dx})_{(2,-2)}=\;\;{1-(2)(-2)^{2}}/{(2)^{2}(-2)+2}=\;\;{1-8}/{-8+2}=\;\;{7}/{6}

Equation of tangent at

(2,-2)

y+2=7/6(x-2)

7x-6y=26

(-2-7)

does not lie on the line

7x-6y=26

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