Vertices, asymptotes for hyperbola using GC

For H2 A level mathematics, JC students seems to have difficulty in memorising the equations of asymptotes. However the GC can calculate them for you (in a way).

Step 1: Go to APPS –> CONICS –> HYPERBOLA

Suppose we want to sketch the graph $\displaystyle \frac{{{\left( x-1 \right)}^{2}}}{{{2}^{2}}}-\frac{{{(y-2)}^{2}}}{3}=1$

Step 2: DO NOT PRESS GRAPH. Instead press “ALPHA” + “ENTER”

The “CENTER” is where the asymptotes intersect and the “SLOPE” is the gradient for each asymptote. By using basic concept of straight line, the asymptotes are $\displaystyle y-2=\pm 1.5(x-1)$.

If the questions are set such that the values are not “nice”, e.g. including surds then the G.C will be unable to give the values in exact form.

Calculating probability: Any difference whether I use P&C or not?

An example:

We have 10 diamond cards and 3 spade cards whereby each of the cards containing the same shape is assigned to a different number. Find the probability that when I draw 3 cards, there are 2 diamonds and 1 spade cards.

Method 1: Using P&C

Probability required = $\displaystyle \frac{{}^{10}{{C}_{2}}\times {}^{3}{{C}_{1}}}{{}^{13}{{C}_{3}}}=\frac{135}{286}$

Method 2: Using probability (similar to tree diagram)

Probability required = $\displaystyle \frac{10}{13}\times \frac{9}{12}\times \frac{3}{11}\times 3=\frac{135}{286}$

A question students may have:

One common question a student may have is: why does Method 2 contain the “$\displaystyle \times 3$” while Method 1 doesn’t? To investigate, let us apply the definition of combination:

$\displaystyle {}^{n}{{C}_{r}}=\frac{n!}{\left( n-r \right)!r!}$

For Method 1:

$\displaystyle \frac{^{10}{{C}_{2}}{{\times }^{3}}{{C}_{1}}}{^{13}{{C}_{3}}}=\left( \frac{10!}{8!2!} \right)\times \left( \frac{3!}{2!1!} \right)\div \left( \frac{13!}{10!3!} \right)$

$\displaystyle =\left( \frac{10\times 9}{2} \right)\times \left( \frac{3}{1} \right)\div \left( \frac{13\times 12\times 11}{1\times 2\times 3} \right)$

$\displaystyle =\frac{10}{13}\times \frac{9}{12}\times \frac{3}{11}\times 3$

which gives us the same answer as Method 2.

In Method 2:We are performing the experiment as if we are drawing 1 card at a time.

The reason why we need to multiply by 3 is because to obtain 2 diamonds and 1 spade, the following events are different even though the end product is the same:

a) diamond, followed by diamond, followed by spade

b) diamond, followed by spade, followed by diamond

$\int{\cos {{(\ln x)}_{{}}}}dx$
where she chose $u=1,dv=\cos (\ln x)$. The reason was she interprets “1” as ${{x}^{0}}$ which is under “A” of the LIATE method. Since “A” is before “T”, she chose to differentiate one instead and got stuck.