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

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Question

(D) Given curve: $x^2y^2 - 2x = 4 - 4y$. Differentiating with respect to $x$: $2xy^2 + 2x^2y \frac{dy}{dx} - 2 = -4 \frac{dy}{dx}$. At point $(2, -2)$

Vedclass · Accepted Answer

$(-2, -7)$

Vedclass · Answer

(D) Given curve: $x^2y^2 - 2x = 4 - 4y$. Differentiating with respect to $x$: $2xy^2 + 2x^2y \frac{dy}{dx} - 2 = -4 \frac{dy}{dx}$. At point $(2, -2)$: $2(2)(-2)^2 + 2(2)^2(-2) \frac{dy}{dx} - 2 = -4 \frac{dy}{dx}$. $16 - 16 \frac{dy}{dx} - 2 = -4 \frac{dy}{dx}$. $14 = 12 \frac{dy}{dx} \Rightarrow \frac{dy}{dx} = \frac{14}{12} = \frac{7}{6}$. Equation of the tangent at $(2, -2)$: $y - (-2) = \frac{7}{6}(x - 2)$. $6(y + 2) = 7(x - 2) \Rightarrow 6y + 12 = 7x - 14 \Rightarrow 7x - 6y = 26$. Checking the options: For $(-2, -7)$: $7(-2) - 6(-7) = -14 + 42 = 28 
eq 26$. Thus,the tangent does not pass through $(-2, -7)$.

The tangent at the point $(2, -2)$ to the curve $x^2y^2 - 2x = 4(1 - y)$ does not pass through the point:

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