# 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

c) spade, followed by diamond, followed by spade.

However using P & C, for combination, we do not need to consider the order of which card is drawn (or should I say the 3 events are already factored in). You can view it as we are taking 3 cards at one go instead of 1 at a time.

# Integration by parts: Always use “LIATE”?

My student was trying out the following question that requires integration by parts: $\int{\cos {{(\ln x)}_{{}}}}dx$

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.

This is one reason why the topic of techniques of integration, especially by parts, requires some foresight. One should consider what happens to the working upon his / her choice for “u”. This example shows us that we should not learn techniques of integration by memorising formulas blindly.