How to Find Equations of tangent lines and Normal Lines:
Practice Problems

$$ f(5) = \sqrt{2(5) - 1} = \sqrt 9 = 3 $$

The point is $$(5, 3)$$.

Since $$f(x) = (2x-1)^{1/2}$$, the derivative is

$$ \begin{align*} f'(x) & = \frac 1 2 (2x-1)^{-1/2}\cdot 2\\[6pt] & = (2x -1)^{-1/2}\\[6pt] & = \frac 1 {\sqrt{2x-1}} \end{align*} $$

$$ f'(5) = \frac 1 {\sqrt{2(5)-1}} = \frac 1 {\sqrt 9} = \frac 1 3 $$

The slope of the tangent line is $$m = \frac 1 3$$.

$$ \begin{align*} y - y_1 & = m(x-x_1)\\[6pt] y - 3 & = \frac 1 3(x-5) \end{align*} $$

The tangent line is $$y - 3 = \frac 1 3(x-5)$$.

For reference, here is the graph of the function and the tangent line we just found.

$$ f(-1) = e^{2(-1)} = e^{-2} \approx 0.3679 $$

The point is $$\left(-1,e^{-2}\right)$$.

$$ f'(x) = 2e^{2x}\longrightarrow f'(-1) = 2e^{-2} \approx 0.7358 $$

The slope of the tangent line at this point is $$m = 2e^{-2}$$.

$$ \begin{align*} y - y_1 & = m(x-x_1)\\[6pt] y - e^{-2} & = 2e^{-2}(x - (-1))\\[6pt] y - e^{-2} & = 2e^{-2}(x + 1) \end{align*}. $$

$$\displaystyle y - e^{-2} = 2e^{-2}(x + 1)$$, which is approximately the line $$y - 0.3679 = 0.7358(x+1)$$

For reference, the graph of the function and the tangent line we just found are shown below.

Since we are given the $$y$$-value, we will need to solve for the $$x$$-value.

$$ \begin{align*} \sqrt[3]{1-3x} & = -2\\[6pt] 1-3x & = (-2)^3\\[6pt] 1 - 3x & = -8\\[6pt] 1 + 8 & = 3x\\[6pt] 9 & = 3x\\[6pt] x & = 3 \end{align*} $$

The point is $$(3,-2)$$.

Since the function is $$f(x) = (1-3x)^{1/3}$$, the derivative is

$$ \begin{align*} f'(x) & = \frac 1 3(1-3x)^{-2/3}\cdot (-3)\\[6pt] & = -(1-3x)^{-2/3}\\[6pt] & = -\frac 1 {(1-3x)^{2/3}}\\[6pt] & = -\frac 1 {\sqrt[3]{(1-3x)^2}} \end{align*} $$

$$ \begin{align*} f'(3) & = -\frac 1 {\sqrt[3]{(1 - 3(3))^2}}\\[6pt] & = -\frac 1 {\sqrt[3]{(-8)^2}}\\[6pt] & = -\frac 1 {\sqrt[3]{64}}\\[6pt] & = -\frac 1 4\\[6pt] & = -\frac 1 4 \end{align*} $$

The slope of the tangent line at this point is $$m = -\frac 1 4$$.

$$ \begin{align*} y - y_1 & = m(x-x_1)\\[6pt] y - (-2) & = -\frac 1 4(x-3)\\[6pt] y + 2 & = -\frac 1 4(x-3) \end{align*} $$

The tangent line at this point is $$y+2 = -\frac 1 4(x-3)$$.

For reference, the graph of the function and the tangent line we found is shown below

Since we are given the function value, we will need to solve for the $$x$$-value.

$$ \begin{align*} \frac{x+1}{x-2} & = 2\\[6pt] x + 1 & = 2(x-2)\\[6pt] x + 1 & = 2x - 4\\[6pt] 1 + 4 & = 2x - x\\[6pt] 5 & = x \end{align*} $$

The point is $$(5,2)$$.

$$ \begin{align*} f'(x) & = \frac{(x-2)(1) - (x+1)(1)}{(x-2)^2}\\[6pt] & = \frac{x - 2 - x - 1}{(x-2)^2}\\[6pt] & = -\frac 3 {(x-2)^2} \end{align*} $$

$$ \begin{align*} f'(5) & = -\frac 3 {(5-2)^2} = -\frac 3 9 = -\frac 1 3 \end{align*} $$

The slope of the tangent line at this point is $$m=-\frac 1 3$$.

$$ \begin{align*} y-y_1 & = m(x-x_1)\\[6pt] y - 2 & = -\frac 1 3(x - 5) \end{align*} $$

$$ y - 2 = -\frac 1 3(x - 5) $$

For reference purposes, the graph of the function and the tangent line we found are shown below.

$$ f'(x) = 2x + 7 $$

$$ \begin{align*} 2x + 7 & = 4\\[6pt] 2x & = -3\\[6pt] x & = -\frac 3 2 \end{align*} $$

$$ \begin{align*} f\left(-\frac 3 2\right) & = \left(-\frac 3 2\right)^2 + 7\left(-\frac 3 2\right) + 5\\[6pt] & = \frac 9 4 - \frac{21} 2 + 5\\[6pt] & = \frac 9 4 - \frac{42} 4 + \frac{20} 4\\[6pt] & = \frac{29 - 42} 4\\[6pt] & = -\frac{13} 4\\[6pt] \end{align*} $$

The point on the function is $$\left(-\frac 3 2, -\frac{13} 4\right)$$.

$$ \begin{align*} y - y_1 & = m(x-x_1)\\[6pt] y - \left(-\frac{13} 4\right) & = 4\left(x - \left(-\frac 3 2\right)\right)\\[6pt] y + \frac{13} 4 & = 4\left(x +\frac 3 2\right) \end{align*} $$

$$ \displaystyle y + \frac{13} 4 = 4\left(x +\frac 3 2\right) $$

For reference, the graph of the function and the tangent line we found is shown below.

$$ f'(x) = \sec^2 x $$

$$ \begin{align*} \sec^2 x & = 1\\[6pt] \sec x & = \pm 1 \end{align*} $$

We know that $$\sec x = -1$$ when $$x = \frac{3\pi} 2 + 2\pi n$$ for $$n = 0, \pm 1, \pm 2,\ldots$$. All of these values are outside the given interval for $$x$$. Consequently, we only need to examine when $$\sec x = 1$$.

$$ \begin{align*} \sec x & = 1\\[6pt] \frac 1 {\cos x} & = 1\\[6pt] \cos x & = 1\\[6pt] x & = \arccos 1\\[6pt] x & = 0 + 2\pi n,\quad n=0,\pm 1,\pm 2,\ldots \end{align*} $$

The only $$x$$-value within the allowed interval is $$x = 0$$.

$$ f(0) = \tan 0 = 0 $$

The point on the function is $$(0,0)$$.

$$ \begin{align*} y - y_1 & = m(x-x_1)\\[6pt] y - 0 & = 1(x-0)\\[6pt] y & = x \end{align*} $$

The equation of the tangent line is $$y = x$$.

For reference, the graph of the function and the tangent line is shown below.

$$ f(3) = 2 + \frac 1 {3^2 + 1} = 2 + \frac 1 {10} = \frac{21}{10} $$

The point on the function is $$\left(3, \frac{21}{10}\right)$$.

Since $$f(x) = 2 + (x^2+1)^{-1}$$, the derivative is

$$ \begin{align*}% f'(x) & = -1(x^2+1)^{-2}\cdot (2x)\\[6pt] & = -\frac{2x}{(x^2+1)^2} \end{align*} $$

$$ \begin{align*} f'(3) & = -\frac{2(3)}{(3^2+1)^2}\\[6pt] & = -\frac 6 {10^2}\\[6pt] & = - \frac 6 {100}\\[6pt] & = - \frac 3 {50} \end{align*} $$

The slope of the tangent line is $$m = -\frac 3 {50}$$.

Since the slope of the tangent line is $$(m = -\frac 3 {50}$$, the slope of the normal line is $$m = \frac{50} 3$$.

$$ \begin{align*} y-y_1 & = m(x-x_1)\\[6pt] y-\frac{21}{10} & = \frac{50} 3(x-3) \end{align*} $$

$$ y-\frac{21}{10} = \frac{50} 3(x-3) $$

For reference, the graph of the function and the normal line are shown below.

$$ f(4) = \ln (4^2 -1) = \ln(15) \approx 2.708 $$

The point on the function is $$(4,\ln 15)$$.

$$ \begin{align*} f'(x) & = \frac 1 {x^2-1}\cdot (2x)\\[6pt] & = \frac{2x}{x^2-1}\\[6pt] & \\[6pt] f'(4) & = \frac{2(4)}{4^2-1}\\[6pt] & = \frac 8 {15} \end{align*} $$

The slope of the tangent line is $$m = \frac 8 {15}$$. This means the slope of the normal line is $$m = -\frac{15} 8$$.

$$ \begin{align*} y-y_1 & = m(x-x_1)\\[6pt] y-\ln 15 & = -\frac{15} 8(x-4) \end{align*} $$

$$ y-\ln 15 = -\frac{15} 8(x-4) $$

For reference, the graph of the function and the normal line are shown below.

$$ \begin{align*} 4e^{-0.5x} & = 8\\[6pt] e^{-0.5x} & = 2\\[6pt] -0.5x & = \ln 2\\[6pt] x & = -2\ln 2\\[6pt] & \approx -1.386 \end{align*} $$

The point of tangency is $$(-2\ln 2, 8)$$.

$$ \begin{align*} f'(x) & = 4e^{-0.5x}\cdot (-0.5)\\[6pt] & = -2e^{-0.5x} \end{align*} $$

Evaluating the derivative at $$x = -2\ln 2$$, we get the following.

$$ \begin{align*} f'(\blue{-2\ln 2}) & = -2e^{-0.5\blue{(-2\ln 2)}}\\[6pt] & = -2e^{\ln 2}\\[6pt] & = -2(2)\\[6pt] & = -4 \end{align*} $$

So the slope of the tangent line at $$x = -2\ln 2$$ is $$m = -4$$. This means the slope of the normal line is $$m = \frac 1 4$$.

$$ \begin{align*} y-y_1 & = m(x-x_1)\\[6pt] y - 8 & = \frac 1 4(x-(-2\ln2))\\[6pt] y - 8 & = \frac 1 4(x+2\ln 2) \end{align*} $$

$$ y-8 = \frac 1 4(x+2\ln 2) $$

For reference, the graph of the function and the normal line is shown below.

$$ \begin{align*} \frac 3 2(x-3)^{2/3} & = 6\\[6pt] (x-3)^{2/3} & = 4\\[6pt] \left((x-3)^{2/3}\right)^3 & = 4^3\\[6pt] (x-3)^2 & = 64\\[6pt] x-3 & = \pm\sqrt{64}\\[6pt] x & = 3\pm 8\\[6pt] \end{align*} $$

Since the question requires $$x\geq 0$$, we conclude $$x = 11$$. This means the point of tangency is $$(11,6)$$.

$$ \begin{align*} f'(x) & = \frac 3 2\cdot \frac 2 3(x-3)^{-1/3}\cdot (1)\\[6pt] & = (x-3)^{-1/3}\\[6pt] & = \frac 1 {\sqrt[3]{x-3}} \end{align*} $$

$$ \begin{align*} f'(11) & = \frac 1 {\sqrt[3]{11-3}}\\[6pt] & = \frac 1 {\sqrt[3] 8}\\[6pt] & = \frac 1 2 \end{align*} $$

The slope of the tangent line is $$m = \frac 1 2$$. This means the slope of the normal line is $$m = -2$$.

$$ \begin{align*} y-y_1 & = m(x-x_1)\\[6pt] y-6 & = -2(x-11) \end{align*} $$

$$ y - 6 = -2(x-11) $$

For reference, the graph of the function and the normal line is shown below.

$$ \begin{align*}% f'(x) & =-\frac 1 4\cdot 2(x+2)\\[6pt] & =-\frac 1 2(x+2)\\[6pt] \end{align*} $$

Since the normal line has a slope of $$m=5$$, the tangent line will have a slope of $$m = -\frac 1 5$$. So, to find the $$x$$-value of the point on the function, we set the derivative equal to $$-\frac 1 5$$ and solve.

$$ \begin{align*} -\frac 1 2(x+2) & = -\frac 1 5\\[6pt] x+2 & = \frac 2 5\\[6pt] x & = \frac 2 5 - 2\\[6pt] & = - \frac 8 5 \end{align*} $$

$$ \begin{align*}% f\left(-\frac 8 5\right) & = -\frac 1 4\left(-\frac 8 5 + 2\right)^2 + 1\\[6pt] & = -\frac 1 4\left(\frac 2 5\right)^2 + 1\\[6pt] & = -\frac 1 4\left(\frac 4 {25}\right) + 1\\[6pt] & = -\frac 1 {25} + \frac{25}{25}\\[6pt] & = \frac{24}{25} \end{align*} $$

So the point of tangency is $$\left(-\frac 8 5, \frac{24}{25}\right)$$.

$$ \begin{align*} y-y_1 & = m(x-x_1)\\[6pt] y-\frac{24}{25} & = 5\left(x-\left(-\frac 8 5\right)\right)\\[6pt] y-\frac{24}{25} & = 5\left(x+\frac 8 5\right) \end{align*} $$

$$ \displaystyle y-\frac{24}{25} = 5\left(x+\frac 8 5\right) $$

For reference, the graph of the function and the normal line we found are shown below.

$$ \begin{align*} f'(x) & = \frac{(3-x)\cdot 2 - (2x+1)(-1)}{(3-x)^2}\\[6pt] & = \frac{6-2x + 2x+1}{(3-x)^2}\\[6pt] & = \frac 7 {(3-x)^2}\\[6pt] \end{align*} $$

Since the normal line has a slope of $$m=-7$$, the tangent line will have a slope of $$m=\frac 1 7$$. So, we set the derivative equal to $$\frac 1 7$$ and solve for $$x$$.

$$ \begin{align*} \frac 7 {(3-x)^2} & = \frac 1 7\\[6pt] \frac {(3-x)^2} 7 & = 7\\[6pt] (3-x)^2 & = 49\\[6pt] 3-x & = \pm 7\\[6pt] 3\pm 7 & = x \end{align*} $$

Since $$x<0$$, we choose $$x = 3 - 7 = -4$$.

$$ \begin{align*} f(-4) & = \frac{2(-4)+1}{3-(-4)}\\[6pt] & = \frac{-8+1}{3+4}\\[6pt] & = \frac{-7}{7}\\[6pt] & = -1 \end{align*} $$

The point of tangency is $$(-4,-1)$$.

$$ \begin{align*} y-y_1 & = m(x-x_1)\\[6pt] y - (-1) & = -7(x-(-4))\\[6pt] y + 1 & = -7(x+4)\\[6pt] \end{align*} $$

$$ y + 1 = -7(x+4) $$

For reference, the graph of the function and the normal line is shown below.

$$ \begin{align*} 16\blue{x^2} + 9y^2 & = 85\\[6pt] 16\blue{\left(\frac 1 2\right)^2} + 9y^2 & = 85\\[6pt] 4 + 9y^2 & = 85\\[6pt] 9y^2 & = 81\\[6pt] y^2 & = 9\\[6pt] y & = \pm 3 \end{align*} $$

Since we are told $$y<0$$, we have the point $$\left(\frac 1 2,-3\right)$$.

Since the ellipse is defined implicitly, we will need to use implicit differentiation.

$$ \begin{align*} 16(2x) + 9\left(2y\,\frac{dy}{dx}\right) & = 0\\[6pt] 32x + 18y\,\frac{dy}{dx} & = 0\\[6pt] 18y\,\frac{dy}{dx} & = -32x\\[6pt] \frac{dy}{dx} & = -\frac{32x}{18y}\\[6pt] & = -\frac{16x}{9y} \end{align*} $$

$$ \begin{align*} \frac{dy}{dx}\bigg|_{\left(\blue{\frac 1 2},\red{-3}\right)} & = -\frac{16\blue{\left(\frac 1 2\right)}}{9\red{\left(-3\right)}}\\[6pt] & = -\frac{8}{-27}\\[6pt] & = \frac{8}{27} \end{align*} $$

The slope of the tangent line at the point of tangency is $$m = \frac 8 {27}$$.

$$ \begin{align*} y - y_1 & = m(x-x_1)\\[6pt] y -\left(-3\right) & = \frac 8 {27}\left(x - \frac 1 2\right)\\[6pt] y + 3 & = \frac 8 {27}\left(x - \frac 1 2\right) \end{align*} $$

$$ \displaystyle y + 3 = \frac 8 {27}\left(x - \frac 1 2\right) $$

For reference, the graph of the ellipse and the tangent line are shown below.

Since the curve is defined implicitly, we will use implicit differentiation. Note that the right-hand side of the equation will require the product rule.

$$ \begin{align*} 3x^2 + 3y^2\,\frac{dy}{dx} & = 9(1)y + 9x\,\frac{dy}{dx}\\[6pt] 3y^2\,\frac{dy}{dx} - 9x\,\frac{dy}{dx} & = 9y - 3x^2\\[6pt] y^2\,\frac{dy}{dx} - 3x\,\frac{dy}{dx} & = 3y - x^2\\[6pt] \frac{dy}{dx}\left(y^2 - 3x\right) & = 3y - x^2\\[6pt] \frac{dy}{dx} & = \frac{3y - x^2}{y^2 - 3x} \end{align*} $$

$$ \begin{align*} \frac{dy}{dx}_{(\blue{2},\red{4})} & = \frac{3\red{(4)} - \blue{(2)}^2}{\red{(4)}^2 - 3\blue{(2)}}\\[6pt] & = \frac{12 - 4}{16 - 6}\\[6pt] & = \frac{8}{10}\\[6pt] & = \frac 4 5 \end{align*} $$

$$ \begin{align*} y-y_1& = m(x-x_1)\\[6pt] y-4 & = \frac 4 5(x-2) \end{align*} $$

$$ y-4 = \frac 4 5(x-2) $$

For reference, the graph of the curve and the tangent line are shown below.

$$ \begin{align*}% (x^2+y^2)(y^2 + x(x+\blue a)) & = 4\blue{a}xy^2\\[6pt] (x^2+y^2)(y^2 + x(x+\blue 2)) & = 4\blue{(2)}xy^2\\[6pt] (x^2+y^2)(y^2 + x(x+2)) & = 8xy^2 \end{align*} $$

$$ \begin{align*} (x^2+y^2)(y^2 + \red x(\red x+2)) & = 8\red{(x)}y^2\\[6pt] (x^2+y^2)(y^2 + \red 1(\red 1+2)) & = 8\red{(1)}y^2\\[6pt] (1+y^2)(y^2 + 3) & = 8y^2\\[6pt] y^2 + 3 + y^4 + 3y^2 & = 8y^2\\[6pt] y^4 + 4y^2 + 3& = 8y^2\\[6pt] y^4 - 4y^2 + 3 & = 0\\[6pt] (y^2 - 3)(y^2-1) & = 0\\[6pt] y^2 = 3 & \quad y^2 = 1\\[6pt] y = \pm\sqrt 3 &\quad y = \pm 1 \end{align*} $$

Since we need positive $$y$$-values, we have the points $$(1,1)$$ and $$(1,\sqrt 3)$$.

Since the trifolium curve is defined implicitly, we will use implicit differentiation. Note that the both sides of the equation involve products, so we will need the product rule.

Before differentiating, we will rewrite the original equation as follows:

$$ \begin{align*} (x^2+y^2)(y^2 + x(x+2)) & = 8xy^2\\[6pt] (x^2+y^2)(y^2 + x^2+2x) & = 8xy^2. \end{align*} $$

Now to differentiate.

$$ \begin{align*} \left(2x+2y\,\frac{dy}{dx}\right)(y^2 + x^2+2x) + (x^2 + y^2)\left(2y\,\frac{dy}{dx} + 2x + 2\right) & = 8y^2 + 16xy\,\frac{dy}{dx} \end{align*} $$

Next, we distribute.

$$ \begin{align*} \left(2x+2y\,\frac{dy}{dx}\right)\blue{(y^2 + x^2+2x)} + \red{(x^2 + y^2)}\left(2y\,\frac{dy}{dx} + 2x + 2\right) & = 8y^2 + 16xy\,\frac{dy}{dx}\\[6pt] 2x\blue{(y^2 + x^2+2x)}+2y\blue{(y^2 + x^2+2x)}\,\frac{dy}{dx} + 2y\red{(x^2 + y^2)}\,\frac{dy}{dx} + (2x + 2)\red{(x^2 + y^2)} & = 8y^2 + 16xy\,\frac{dy}{dx} \end{align*} $$

Collect the $$\frac{dy}{dx}$$ terms on the left, factor, then solve.

$$ \begin{align*} 2y(y^2 + x^2+2x)\,\frac{dy}{dx} + 2y(x^2 + y^2)\,\frac{dy}{dx} - 16xy\,\frac{dy}{dx}& = 8y^2 - 2x(y^2 + x^2+2x) - (2x + 2)(x^2 + y^2)\\[6pt] \frac{dy}{dx}\left[2y(y^2 + x^2+2x) + 2y(x^2 + y^2) - 16xy\right] & = 8y^2 - 2x(y^2 + x^2+2x) - (2x + 2)(x^2 + y^2)\\[6pt] \frac{dy}{dx} & = \frac{8y^2 - 2x(y^2 + x^2+2x) - (2x + 2)(x^2 + y^2)}{2y(y^2 + x^2+2x) + 2y(x^2 + y^2) - 16xy} \end{align*} $$

Simplify by expanding the numerator and denominator and collecting like terms.

$$ \begin{align*} \frac{dy}{dx} & = \frac{8y^2 - 2xy^2 - 2x^3 - 4x^2 -2x^3 - 2xy^2 - 2x^2 - 2y^2}{2y^3 + 2x^2y+4xy + 2x^2y + 2y^3 - 16xy}\\[6pt] \frac{dy}{dx} & = \frac{6y^2 - 4xy^2 - 4x^3 - 6x^2}{4y^3 + 4x^2y - 12xy}\\[6pt] \frac{dy}{dx} & = \frac{2(3y^2 - 2xy^2 - 2x^3 - 3x^2)}{2(2y^3 + 2x^2y - 6xy)}\\[6pt] \frac{dy}{dx} & = \frac{3y^2 - 2xy^2 - 2x^3 - 3x^2}{2y^3 + 2x^2y - 6xy} \end{align*} $$

At $$(1,1)$$ the slope of the tangent line is

$$ \begin{align*} \frac{dy}{dx}\bigg|_{(\blue{1},\red{1})} & = \frac{3\red{(1)}^2 - 2\blue{(1)}\red{(1)}^2 - 2\blue{(1)}^3 - 3\blue{(1)}^2}{2\red{(1)}^3 + 2\blue{(1)}^2\red{(1)} - 6\blue{(1)}\red{(1)}}\\[6pt] & = \frac{3 - 2 -2 - 3}{2 + 2 - 6}\\[6pt] & = \frac{-4}{-2}\\[6pt] & = 2. \end{align*} $$

At $$(1,\sqrt 3)$$, the slope of the tangent line is

$$ \begin{align*} \frac{dy}{dx}\bigg|_{(\blue{1},\red{\sqrt 3})} & = \frac{3\red{(\sqrt 3)}^2 - 2\blue{(1)}\red{(\sqrt 3)}^2 - 2\blue{(1)}^3 - 3\blue{(1)}^2}{2\red{(\sqrt 3)}^3 + 2\blue{(1)}^2\red{(\sqrt 3)} - 6\blue{(1)}\red{(\sqrt 3)}}\\[6pt] & = \frac{3(3) - 2(3) - 2 - 3}{2(3\sqrt 3) + 2\sqrt 3 - 6\sqrt 3}\\[6pt] & = \frac{-2}{2\sqrt 3}\\[6pt] & = - \frac 1 {\sqrt 3}\\[6pt] & = -\frac{\sqrt 3} 3 \end{align*} $$

At $$(1,1)$$ the slope of the normal line is $$m = -\frac 1 2$$.

At $$(1,\sqrt 3)$$ the slope of the normal line is $$m = \sqrt 3$$.

At $$(1,1)$$ the equation of the normal line is

$$ y - 1 = -\frac 1 2(x-1) $$

At $$(1,\sqrt 3)$$ the equation of the normal line is

$$ y - \sqrt 3 = \sqrt 3(x-1) $$

The equations of the normal lines are $$y - 1 = -\frac 1 2(x-1)$$ and $$y-\sqrt 3 = \sqrt 3(x-1)$$.

For reference, the graph of the trifolium and the two normal lines is shown below.

$$ \begin{align*} y^2 & = 9(3-\blue{2})^5\,\blue{(2)}^2\\[6pt] y^2 & = 9(1)^5\,(4)\\[6pt] y^2 & = 36\\[6pt] y & = \pm\sqrt{36} \end{align*} $$

Since we want $$y>0$$, the point of tangency is $$\left(2, 6\right)$$.

Note that the equation is implicitly defined, so we will use implicit differentiation. Additionally, the right-hand side of the equation requires the product rule.

$$ \begin{align*} 2y\,\frac{dy}{dx} & = 9(5)(3-x)^4(-1)\cdot x^2 + 9(3-x)^5\cdot 2x\\[6pt] 2y\,\frac{dy}{dx} & = 9x\left[-5x(3-x)^4 + 2(3-x)^5\right]\\[6pt] 2y\,\frac{dy}{dx} & = 9x(3-x)^4\left[-5x + 2(3-x)\right]\\[6pt] 2y\,\frac{dy}{dx} & = 9x(3-x)^4(-5x + 6-2x)\\[6pt] 2y\,\frac{dy}{dx} & = 9x(3-x)^4(6-7x)\\[6pt] \frac{dy}{dx} & = \frac{9x(6-7x)(3-x)^4}{2y} \end{align*} $$

$$ \begin{align*} \frac{dy}{dx}\bigg|_{\left(\blue{2}, \red{6}\right)} & = \frac{9\blue{(2)}(6-7\blue{(2)})(3-\blue{(2)})^4}{2\red{(4)}}\\[6pt] & = \frac{9(6-14)(1)^4} 4\\[6pt] & = \frac{9(-8)} 4\\[6pt] & = -18 \end{align*} $$

The slope of the tangent line is $$m = -18$$, so the slope of the normal line is $$m = \frac 1 {18}$$.

$$ \begin{align*} y-y_1 & = m(x-x_1)\\[6pt] y-6 & = \frac 1 {18}(x-2) \end{align*} $$

$$ \displaystyle y-6 = \frac 1 {18}(x-2) $$

For reference, the curve and the normal line are shown below.