### JEE Main On 8 April 2017 Question 9

##### Question: The tangent at the point (2,-2) to the curve $ x^{2}y^{2}-2x=4(1-y) $ does not pass through the point: [JEE Online 08-04-2017]

#### Options:

A) (-2,-7)

B) (8, 5)

C) (-4,-9)

D) $ ( 4,\frac{1}{3} ) $

## Show Answer

#### Answer:

Correct Answer: A

#### Solution:

$ x^{2}y^{2}-2x=4-4y $

$ 2xy^{2}+2y.x^{2}.\frac{dy}{dx}-2=-4.\frac{dy}{dx} $

$ \frac{dy}{dx}(2y.x^{2}+4)=2-2x.y^{2} $ $ {{. \frac{dy}{dx} |} _{2,-2}} $

$ =\frac{2-2\times 2\times 4}{2(-2)\times 4+4}=\frac{+14}{+12}=\frac{7}{6} $

$ (y+2)=\frac{7}{6}(x-2)\Rightarrow 7x-6y=26 $

(−2,−7) does not satisfy above eq.