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.

• A. $(8,5)$
• B. $(4,\frac{1}{3})$
• C. $(-2,-7)$
• D. $(-4,-9)$

Answer 1

Answer: (C)

$(-2,-7)$

$x^2{y}^2-2x=4-4y$
$2x{y}^2+2y\cdot x^2\cdot \frac{dy}{dx}-2=-4\cdot \frac{dy}{dx}$
$\frac{dy}{dx}(2y\cdot x^2+4)=2-2x\cdot y^2$
${\frac{dy}{dx}|}_{2,-2}=\frac{2-2\times 2\times 4}{2(-2)\times 4+4}=\frac{+14}{+12}=\frac{7}{6}$
eq of tangent :$(y+2)=\frac{7}{6}(x-2)\Rightarrow 7x-6y=26$.

$(-2,-7)$ does not satisfy above eq.