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Use implicit differentiation, treating $y$ as a function of $x$:Where $y\ne 0$, this is $\boxed{dy/dx=-2x/(3y)}$. At the two points where $y=0$, the ellipse has vertical tangent lines rather than a finite slope.

The two derivatives are $2x$ and $2(x-1)$. At the intersection their slopes are $1$ and $-1$, so their direction vectors $(1,1)$ and $(1,-1)$ have zero dot product. Thus the tangent lines are perpendicular and their angle is

At $x=3$, the point on the parabola is $(3,8)$ and the tangent line has slope $f^{\prime }(3)=6$. The normal line to a plane curve therefore has slope $-1/6$. Its equation isThus the requested results are

The tangent has poles at $\pi /2$ and $3\pi /2$, so these are vertical asymptotes of $x\tan x$. The zeros on the given interval are $0,\pi ,2\pi$. Near zero, $x\tan x=x^2+O(x^4)$, so the graph starts with a horizontal tangent line.

Each of its three branches is increasing. Indeedand the numerator starts at zero with derivative $2{\cos }^{2}x\ge 0$, so it is positive for $x>0$. The first branch rises from zero to $+\infty$, the middle branch rises from $-\infty$ through $(\pi ,0)$ to $+\infty$, and the last rises from $-\infty$ to $(2\pi ,0)$.

The cycloid has limiting endpoints $(0,0)$ and $(2\pi a,0)$, with its highest point $(\pi a,2a)$ at $t=\pi$. The transformation $t\mapsto 2\pi -t$ sends $x$ to $2\pi a-x$ and preserves $y$, so the arch is symmetric about $x=\pi a$. Its slope decreases from $+\infty$ through zero to $-\infty$, giving vertical limiting tangent lines at the cusps and a horizontal tangent line at the top. More explicitly,so the arch is a concave function throughout its interior. The endpoint cusps are limiting points, excluded by the open parameter interval.

The original PDF specifies slope $\sqrt{3}$; the transcription's $\sqrt{-1}$ is erroneous. By part (a), $\cot (t/2)=\sqrt{3}$. Since $t/2\in (0,\pi )$ and cotangent is strictly decreasing there, the unique parameter value is $t=\pi /3$. ThereforeThe tangent line drawn in the preceding figure has the required positive slope.

At a single point of contact the line is a tangent line, so its slope equals the derivative of the exponential function. At the contact coordinate $x_0$,Since $m>0$, dividing gives $x_0=1$, hence $m=e$. To verify uniqueness, the difference $e^x-ex$ has positive second derivative everywhere and its unique minimum is zero at $x=1$. Thus this line touches the curve once.

For this parametric curve, the chain rule givesThe quotient is a finite slope when $\sin t\ne 0$. At $t=0$ and $t=\pi$, the horizontal component of the tangent vanishes while the vertical component does not; these have vertical tangent lines, rather than missing finite values of the displayed quotient.

The normal line to a plane curve is perpendicular to the tangent line, so its slope is the negative reciprocal $1/\sqrt{6}$. The required equation isThe normal direction also agrees with the gradient $(4x,2y)$ of the defining level function.

The Newton root-finding iteration follows the current tangent line to its horizontal-axis intersection. Since $f^{\prime }(x)=2x$, it givesA positive initial guess is appropriate for the positive square root; negative starting values instead target the negative root.

The seven requested parabolas are $y=c{x}^2$ for $c=-3,-2,-1,0,1,2,3$; the case $c=0$ is the horizontal axis. Positive $c$ opens upward, negative $c$ downward, and larger $|c|$ gives a narrower parabola. All pass through the origin with a horizontal tangent line.

To find orthogonal trajectories, use the continuous family $y=c{x}^2$ containing these seven representative members. Its slope is $y^{\prime }=2cx=2y/x$ away from $x=0$. Two finite slopes of perpendicular tangents multiply to $-1$, so the orthogonal slope isThis separable differential equation gives $2ydy=-xdx$, henceThese ellipses are the regular real trajectories; $C=0$ is a single point, not a regular curve, and $C<0$ has no real points. One example is $x^2+2y^2=2$, with semiaxes $\sqrt{2}$ horizontally and $1$ vertically.

The implicit tangent relation $2xdx+4ydy=0$ also describes points with vertical tangents, where the slope quotient is unavailable. At intersections with the member $y=0$, the ellipse has a vertical tangent line and is perpendicular to that horizontal axis. The common origin of the parabolas is not on any regular trajectory with $C>0$. The sketch shows the requested seven members and the example ellipse.

A tangent line through $(1,1)$ with slope $-2/3$ has equation

The two derivatives are $3x^2+3$ and $6x$, both equal to six at $x=1$. Their common tangent line through $(1,4)$ is $y-4=6(x-1)$, or $y=6x-2$. Setting $x=0$ gives its intersection with the vertical axis:

At a point of tangency, the derivative of the curve equals the slope one of the straight line. The chain rule gives $2\cos 2x=1$. Since $0\le 2x\le \pi$, the unique solution is $x=\pi /6$. The corresponding height is $\sin (\pi /3)=\sqrt{3}/2$. Substitution in the tangent line gives

The normal vector direction is orthogonal to the tangent line. For finite nonzero slopes, their product is $-1$. The tangent has slope one, so the normal line has

By implicit differentiation of the ellipse equation,The tangent line through $(1,1)$ is thereforeIts normal vector is $(9,16)$, parallel to the gradient $(18,32)$ of $9x^2+16y^2$ at that point.

Setting the tangent line approximation equal to zero gives the Newton root-finding iterationThe step requires a nonzero derivative at the starting point.

The derivative is $y^{\prime }=2x+1/x$, so the tangent line has slope $3$ at $x=1$. The corresponding point is $(1,1)$ because $\ln 1=0$. Hence the tangent line is $y-1=3(x-1)$, or $y=3x-2$. Thus

The chord through the endpoints is the straight line $y=(x-1)\log 2$. At $\xi =1/\log 2$, the tangent line isThese two straight lines have the same slope, as required by the mean value theorem. Since $f^{\prime \prime }(x)=-1/x^2<0$, the natural logarithm is a concave function: its graph lies above the chord and below this tangent line.

The Newton root-finding iteration replaces the graph by its tangent line at $x_0$ and uses that line's zero:Comparing with the printed reciprocal-ratio form givesThis step requires $f^{\prime }(x_0)\ne 0$. If $f(x_0)=0$, the iteration has already reached a zero of a function and the printed reciprocal-ratio expression need not be evaluated.