The tangent at the point (2, −2) to the curve, x2y2−2x=4(1−y) does not pass through the point :  
Option: 1 (8,5)
Option: 2 (4,1/3)
Option: 3 (-2,-7)
Option: 4 (-4,-9)
 

Answers (1)

As learnt in concept ABCD

 

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

Differentiate both sides wrt.x.

2xy^{y}+x^{2}.2yy{}'-2=4-4y{}'

y{}'=-\frac{(xy^{2}-1)}{2+x^{2}y}=\left ( \frac{1-xy^{2}}{2+x^{2}y} \right )

at (2,-2) y{}'=\frac{1-2\times (-2)^{2}}{2+2^{2}\times (-2)}=\frac{1-8}{2-8}=+\frac{7}{6}

Equation of tangent is

\frac{y+2}{x-2}=\frac{7}{6}=>7x-6y-26=0

It doesn't pass through (-2,-7)

Concept ABCD

Slope of curve at a given point

To find slope, we differentiate with respect to x and find \frac{dy}{dx}

Eg in x^{2}-y^{2}=4

\frac{dy}{dx}=\frac{x}{y}

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