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

Answer 1

The correct option is A $(-2,-7)$

$\begin{array}{c}Wehave\\ Pointof\tan gent=\left(2,2\right)\\ cure:x^2{y}^2-2x=4\left(1-y\right)\\ calculating\frac{dy}{dx}at\left(2,2\right)\\ 2x{y}^2+2x^{2}y\frac{dy}{dx}-2=-4\frac{dy}{dx}\\ 16-16\frac{dy}{dx}-2=-4\frac{dy}{dx}\\ 14=12\frac{dy}{dx}\\ \therefore \frac{dy}{dx}=\frac{7}{6}\\ \tan gent:\\ y+2=\frac{7}{6}\left(x-2\right)\\ Hence,\left(-2,-7\right)dojotsatisfy\\ Hence,theoptionAisthecorectanswer.\end{array}$