Find the tangent line of ellipse without calculus

December 19, 2023December 25, 2023 / Michael Xue / Leave a comment

Suppose an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ has a tangent line $y = kx + m$ at $(x_0, y_0)$ :

$\frac{x_0^2}{a^2} + \frac{y_0^2}{b^2} = 1,\quad\quad\quad(1)$

$y_0 = kx_0 + m.\quad\quad\quad(2)$

Substituting (2) into (1) gives

$\frac{x_0^2}{a^2} + \frac{(kx_0+m)^2}{b^2} = 1,$

a quadratic equation.

Since $(x_0, y_0)$ is the only point of intersection, this quadratic equation has one root. i.e., its discriminant must be zero. As a result,

$x_0 = \frac{\frac{-2 km}{b^2}}{2\left(\frac{k^2}{b^2}+\frac{1}{a^2}\right)} \overset{(2)}{=}\frac{-k(y_0-kx_0)}{b^2 \left(\frac{k^2}{b^2}+\frac{1}{a^2}\right)}.\quad\quad\quad(3)$

Solving (3), we obtain:

$\forall y_0>0, k = \frac{-b^2x_0}{a^2y_0}.$

It follows that $m=y_0+\frac{b^2x_0^2}{a^2y_0}.$

Fig. 1

There is an alternative:

Fig. 2

Exercise-1 Explain the alternative way of obtaining $k$ depicted in Fig. 2. (hint: $a^2y_0^2+b^2x_0^2-a^2b^2=0$ )

Exercise-2 Show that the tangent line can be expressed as $\frac{x_0x}{a^2} + \frac{y_0y}{b^2} = 1$ .

A Proof

December 14, 2023December 17, 2023 / Michael Xue / Leave a comment

Given

where $L$ is the tangent line at $(x_0, y_0)$ and $DB || AC.$ Show that $\alpha = \beta.$

The differentiation of $y^2=4ax$ yields

$2y\cdot y' = 4a \implies$ for $y \ne 0, y' = \frac{2a}{y}.$

It means the slope of the tangent line at $(x_0, y_0)$ is $\frac{2a}{y_0}.$

The equation of the tangent line $L$ then becomes

$y = \frac{2a}{y_0} x + m$ .

When $x=x_0,$

$y_0 = \frac{2a}{y_0}x_0 + m \implies m = y_0 - \frac{2a}{y_0}x_0.$

Substituting this back into the original equation, we have:

$y = \frac{2a}{y_0} x + y_0-\frac{2a}{y_0}x_0.\quad\quad\quad(1)$

At $C$ where $y_c=0$ ,

$x_c\overset{(1)}{=}(\frac{2a}{y_0}x_0- y_0)\frac{y_0}{2a}$

$= \frac{2ax_0-y_0^2}{y_0}\cdot \frac{y_0}{2a}$

$= (2ax_0-4ax_0)\cdot \frac{y_0}{2a}$

$= -2a x_0 \cdot \frac{1}{2a}$

$= -x_0$ .

$AC = \sqrt{(a-(-x_0))^2}$

$= \sqrt{(a+x_0)^2}$

$= |a+x_0|$ .

$AB = \sqrt{(a-x_0)^2+(0-y_0)^2}$

$= \sqrt{a^2-2ax_0+x_0^2+4ax_0}$

$= \sqrt{a^2+2ax_0+x_0^2}$

$= \sqrt{(x_0 + a)^2}$

$= |a+x_0|$ .

Therefore,

$AB= AC \implies \Delta ABC$ is an isosceles triangle $\implies \beta = \gamma.$

Since

$DB || AC \implies \alpha = \gamma,$

it follows that

$\alpha = \beta.$

Exercise-1 Find the tangent line of parabola $y^2=4ax$ without calculus (hint: “Constructing the tangent line of quadratic without calculus“)

Doveryai no Proveryai

December 7, 2023February 18, 2024 / Michael Xue / Leave a comment

Using the Omega CAS Explorer, we can confirm the solution(s) for a given differential equation.

Fig. 1

In Fig. 1, for instance, applying ‘ode2’ to the following differential equation:

$y'' + 5y'+ 6y =\sin(x)e^{-3x},\quad\quad\quad(1)$

yields an explicit expression for $y$ :

$y = \frac{ -e^{-3x}(\sin(x)-cos(x))}{2} + C_1 \cdot e^{-2x} + C_2 \cdot e^{-3x}.\quad\quad\quad(2)$

To validate $y$ as a solution, we substitute (2) into (1) for evaluation. The simplified result confirms that $y$ expressed in (2) indeed satisfies the given differential equation.

Now, consider

$(3y^2+e^x)y' + e^xy+ \cos(x)+e^{x}=0.\quad\quad\quad(3)$

Fig. 2

The solver gives an implicit solution

$y^3+e^xy+\sin(x)+e^x=C.\quad\quad\quad(4)$

Obtaining an explicit $y$ from (4) is challenging. However, the verification is surprisingly simple: By declaring that $y$ is a function of $x$ and then differentiating (4), we recover the solved differential equation! This is an indication that implicit solution (4) satisfies the differential equation. i.e., (4) is a solution of (3).

Fig. 3

Alternatively, the verification process can be simplified by employing the ‘ode_check’ function (see Fig. 3 ) in the ‘contrib_ode’ package, as illustrated in Fig. 4.

Fig. 4

It is noteworthy that while the differential equation is solved using ‘ode2’, if ‘contrib_ode’ is chosen instead, the solution for verification needs to be extracted from a list before invoking ‘ode_check’:

Fig. 5

“Austrian trains are always late. A Prussian visitor asks the Austrian conductor why they bother to print timetables. The conductor replies “If we did not, how would we know how late the trains are?” – Victor Weisskopf

Exercise-1 Obtain (4) without using a CAS.

Exercise-2 For grins, let us solve (4) for the $y$ s and show that they are the solutions of (3).

TriDet Mastery: Mnemonics for Effortless 3×3 Determinant Calculation

December 5, 2023December 5, 2023 / Michael Xue / Leave a comment

I find that the most straightfoward method for calculating a 3 by 3 determinant is to recall the pattern depicted below:

For example, determinant

is calculated as follows:

$=1\cdot(-3)\cdot(-4) + 2\cdot 0\cdot 0 +(-4)\cdot 2 \cdot 1 - 0\cdot(-3)\cdot(-4) -1\cdot 0\cdot 1 - (-4)\cdot 2\cdot 2$

$= 12 + 0 - 8 -0 -0 + 16$

$= 20$ .

Excerise-1 find a mnemonics pattern for complex multiplication $(a+b\bold{i})\cdot(c+d\bold{i})$ .

Excerise-2 Find a mnemonics pattern for complex division $\frac{a+b\bold{i}}{c+d\bold{i}}$ .

V is for Vector

December 4, 2023December 4, 2023 / Michael Xue / Leave a comment

Maxima’s vector is implemented as a list. For example, v: [1,2].

How a vector in Maxima behaves depends on the context it is in.

v:[1,2];
m:matrix([1,2], [3,4]);
m.v;

outputs:

so v is a column vector:

$\begin{pmatrix} 1 \quad 2 \\ 3 \quad 4 \end{pmatrix}\cdot \begin{pmatrix} 1 \\ 2 \end{pmatrix} =\begin{pmatrix} 1\cdot 1 + 2\cdot 2 \\ 3\cdot 1 + 4\cdot 2\end{pmatrix} =\begin{pmatrix} 5 \\ 11 \end{pmatrix}$ .

However, given

v:[1,2];
m:matrix([1,2], [3, 4]);
v.m;

v behaves as a row vector:

i.e.,

$\begin{pmatrix} 1 \quad 2\end{pmatrix} \cdot \begin{pmatrix} 1 \quad 2\\3 \quad 4 \end{pmatrix} = \begin{pmatrix} 1\cdot 1 + 2\cdot 3 \\ 1\cdot 2+ 2\cdot 4 \end{pmatrix} =\begin{pmatrix} 7 \quad10\end{pmatrix}$ .

Maxima is smart enough to compute the inner product of two vectors:

That is,

$\begin{pmatrix}1\quad 2\end{pmatrix}\cdot\begin{pmatrix}3\\4\end{pmatrix}=1\cdot 3 + 2 \cdot 4 = 11$ .

Notice we multiply vectors by ‘.’ . If ‘*’ is used instead then the result is a list which is incorrect:

In the past, I wrote the row vector as a 1 by n matrix and the column vector n by 1 matrix. Thanks to Josef Leydold and Martin Petry, the authors of “Introduction to Maxima for Economics“, I now express the vector as a list and let maxima figure out whether the list is a row vector or column vector.

Piece of Cake

December 1, 2023January 8, 2024 / Michael Xue / Leave a comment

Prove Varignon’s theorem:

The midpoints of the sides of any quadrilateral form the vertices of a parallelogram.

Fig. 1

The quadrilaterals are depicted in Fig. 1. $E, F, G$ and $H$ are the midpoints of the sides.

In a rectangular coordinate system:

For points $A(x_1, y_1), B(x_2, y_2), C(x_3, y_3)$ and $D(x_4, y_4)$ , the midpoints are:

$E(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}), F(\frac{x_2+x_3}{2}, \frac{y_2+y_3}{2}), G(\frac{x_3+x_4}{2}, \frac{y_3+y_4}{2}), H(\frac{x_1+x_4}{2}, \frac{y_1+y_4}{2}).$