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Solution

Differentiate: 

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Solution

Tangent ∥ x-axis ⇒ slope = 0 

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Solution

Curve: \(x^2y^2-2x=4(1-y)\). Point: \((2,-2)\).

Differentiate implicitly:
\(\dfrac{d}{dx}(x^2y^2)-2=\dfrac{d}{dx}\big(4(1-y)\big)\)
\(2xy^2+2x^2y\,y'-2=-4y'\)
\(\Rightarrow y'\,(2x^2y+4)=2-2xy^2\)
\(\Rightarrow y'=\dfrac{1-xy^2}{x^2y+2}\).

Slope at \((2,-2)\):
\(xy^2=2\cdot4=8\), \(x^2y+2=4\cdot(-2)+2=-6\).
\(m=y'=\dfrac{1-8}{-6}=\dfrac{-7}{-6}=\dfrac{7}{6}\).

Tangent line at \((2,-2)\):
\(y+2=\dfrac{7}{6}(x-2)\ \Rightarrow\ 6y=7x-26\ \Rightarrow\ \boxed{\,y=\tfrac{7}{6}x-\tfrac{13}{3}\,}\).

How to decide “does not pass through”: A point \((x_0,y_0)\) lies on the tangent iff \(6y_0=7x_0-26\). If this fails, the tangent does not pass through that point.

Checks (examples): On the line: \((0,-\tfrac{13}{3})\), \((\tfrac{26}{7},0)\). Any point not satisfying \(6y=7x-26\) is not on the tangent.

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Solution

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Solution

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Solution

We have the parametric curve: $$ x = t^2 + 3t - 8,\qquad y = 2t^2 - 2t - 5. $$ The point $(2, -1)$ lies on the curve. Find the value of $t$: Solve $$ t^2 + 3t - 8 = 2. $$ So, $$ t^2 + 3t - 10 = 0. $$ Factor: $$ (t+5)(t-2)=0 \quad\Rightarrow\quad t=2 \text{ or } t=-5. $$ Check which gives $y=-1$: For $t=2$: $$ y = 2(2)^2 - 2(2) - 5 = 8 - 4 - 5 = -1. $$ So the correct parameter is: $$ t = 2. $$ Now compute slope of tangent: $$ \frac{dy}{dx} = \frac{dy/dt}{dx/dt}. $$ Compute derivatives: $$ \frac{dx}{dt} = 2t + 3,\qquad \frac{dy}{dt} = 4t - 2. $$ Thus: $$ \frac{dy}{dx} = \frac{4t - 2}{2t + 3}. $$ Substitute $t=2$: $$ \frac{dy}{dx} = \frac{4(2) - 2}{2(2) + 3} = \frac{8 - 2}{4 + 3} = \frac{6}{7}. $$ Slope of normal $m_N$ is negative reciprocal of tangent slope: $$ m_N = -\frac{1}{\frac{6}{7}} = -\frac{7}{6}. $$ Final Answer: $\displaystyle -\frac{7}{6}$