Solve 45 - 50 in two pages and compress the solutions.

From question 44, the derivative of $y = \frac{x}{x-2}$ is $\frac{dy}{dx} = \frac{-2}{(x-2)^2}$.
At the point $(4, 2)$, the slope of the tangent $m_{tangent}$ is:

The normal line is perpendicular to the tangent line, so $m_{normal} = -\frac{1}{m_{tangent}}$.

Using the point $(x_1, y_1) = (4, 2)$ and the slope $m_{normal} = 2$:

The equation of the normal is:

Given $y = \frac{x^2}{x-1}$.

Substitute $x=2$ into the derivative:

Using the point $(x_1, y_1) = (2, 4)$ and the slope $m = 0$:

The equation of the tangent is:

From question 46, the slope of the tangent $m_{tangent}$ at $(2, 4)$ is 0.

Since the tangent is a horizontal line ($y=4$), the normal line will be a vertical line with an undefined slope.

A vertical line passing through the point $(2, 4)$ has the equation $x = x_1$.

The equation of the normal is:

Given $y = \frac{x^2+1}{x}$. Rewrite as $y = x + x^{-1}$.
Differentiate $y$ with respect to $x$:

Substitute $x=1$ into the derivative:

Using the point $(x_1, y_1) = (1, 2)$ and the slope $m = 0$:

The equation of the tangent is:

From question 48, the slope of the tangent $m_{tangent}$ at $(1, 2)$ is 0.

Since the tangent is a horizontal line ($y=2$), the normal line will be a vertical line with an undefined slope.

A vertical line passing through the point $(1, 2)$ has the equation $x = x_1$.

The equation of the normal is:

Given $y = \sqrt{x}$. Rewrite as $y = x^{1/2}$.
Differentiate $y$ with respect to $x$:

Substitute $x=4$ into the derivative:

Using the point $(x_1, y_1) = (4, 2)$ and the slope $m = \frac{1}{4}$:

The equation of the tangent is: