Author Topic: VCE Specialist 3/4 Question Thread! (Read 2171631 times) Tweet Share

pi · « **Reply #975 on:** December 08, 2012, 01:38:25 pm »

Quote from: Stick on December 07, 2012, 05:48:16 pm

I won't tell you how to do it exactly, but for hyperbolae with the general form $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ , the asymptotes are given by $y=\pm\frac{bx}{a}$ .

Easier way is if you have $\pm \frac{(x-h)^2}{a^2} \mp \frac{(y-k)^2}{b^2} = 1$ , just solve for $y$ using the equation $\pm \frac{(x-h)^2}{a^2} \mp \frac{(y-k)^2}{b^2}=0$

Saves you remembering asymptote formulas that you might muddle up during exam pressures.

Hancock · « **Reply #976 on:** December 08, 2012, 08:32:16 pm »

Dunno Pi, I always just remembered that asymptotes have the equation:

(y-k) = (+ or -)(b/a)(x-h)

Stick · « **Reply #977 on:** December 08, 2012, 08:55:56 pm »

It's not that hard to remember, unlike the general circle formula stuff.

Lasercookie · « **Reply #978 on:** December 08, 2012, 08:57:16 pm »

Quote from: Stick on December 08, 2012, 08:55:56 pm

general circle formula stuff.

What's that? I'm drawing a mind blank on what you mean

Stick · « **Reply #979 on:** December 08, 2012, 09:00:07 pm »

When a circle equation is in the form $ax^2+by^2+Dx+Ey+F=0$ , there are some generic formulae which give you to the centre and radius of the circle without having to complete the square.

Planck's constant · « **Reply #980 on:** December 08, 2012, 09:11:00 pm »

Quote from: Hancock on December 08, 2012, 08:32:16 pm

Dunno Pi, I always just remembered that asymptotes have the equation:

(y-k) = (+ or -)(b/a)(x-h)

Good try, but you don't look anything like the 'remembering' type

Planck's constant · « **Reply #981 on:** December 08, 2012, 09:15:05 pm »

Quote from: Stick on December 08, 2012, 09:00:07 pm

When a circle equation is in the form $ax^2+by^2+Dx+Ey+F=0$ , there are some generic formulae which give you to the centre and radius of the circle without having to complete the square.

This will never be a circle equation unless a=b

Planck's constant · « **Reply #982 on:** December 08, 2012, 09:32:21 pm »

There danger with trying to remember too many formulas in maths is that you don't develop the critical thinking that comes with deriving things.

Stick · « **Reply #983 on:** December 08, 2012, 10:33:19 pm »

Quote from: argonaut on December 08, 2012, 09:15:05 pm

This will never be a circle equation unless a=b

Well of course.

Homer · « **Reply #984 on:** December 10, 2012, 10:35:13 pm »

The equation $x^{2}+ax+y^{2}+1=0,$ , where a is a real constant, will represent a circle if? I'm not sure how to approach this question lol :/

b^3 · « **Reply #985 on:** December 10, 2012, 10:40:28 pm »

Firstly, complete the square so that we can get it into the form for the equation of a circle, $x^{2}+y^{2}=r^{2}$

Spoiler

$\begin{alignedat}{1}x^{2}+ax+y^{2}+1 & =0 \\ x^{2}+ax+\left(\frac{a}{2}\right)^{2}-\left(\frac{a}{2}\right)^{2}+y^{2} & =-1 \\ \left(x+\frac{a}{2}\right)^{2}+y^{2} & =\frac{a^{2}-4}{4} \end{alignedat}$

Now for this to be a circle, the RHS of the equation cannot be $0$ and has to be positive, otherwise we would have a radius of $0$ and well, we would have no circle.

Spoiler

$\begin{alignedat}{1}\frac{a^{2}-4}{4} & >0 \\ a^{2} & >4 \\ a<-2 & \: or\: a>2 \\ a & \in R\backslash\left[-2,2\right] \end{alignedat} $

EDIT: Stuffed, up fixed it now.

Homer · « **Reply #986 on:** December 10, 2012, 10:41:27 pm »

Also, $P$ is any point on the hyperbola with equation $x^{2}-\frac{y^{2}}{4}=1$ .

If $m$ is the gradient of the hyperbola at $P$ , then $m$ could be:

polar · « **Reply #987 on:** December 10, 2012, 11:04:23 pm »

using implicit differentiation
$\begin{aligned} x^2 - \frac{y^2}{4} &= 1 \\ \frac{d}{dx}(x^2-\frac{y^2}{4}) &= \frac{d}{dx}(1) \\ \frac{d}{dx}(x^2) - \frac{d}{dx}(\frac{y^2}{4}) &= 0 \\ 2x - \frac{d}{dx}(\frac{y^2}{4}) &= 0 \end{aligned}$

but we need to differentiate with respect to x in the end and not y, so we introduce a new variable and apply the chain rule

$\begin{aligned}\\ \text{Let\;} u=\frac{y^2}{4}, \; \frac{du}{dy} &= \frac{y}{2} \\ \frac{du}{dx} &= \frac{du}{dy}\frac{dy}{dx} \\ \frac{du}{dy}\frac{dy}{dx} &= \frac{y}{2}\frac{dy}{dx} \\ 2x - \frac{du}{dx} &= 0 \\ 2x - \frac{y}{2}\frac{dy}{dx} &= 0 \\ \frac{y}{2}\frac{dy}{dx} &= 2x \\ \frac{dy}{dx} &= \frac{2x}{\frac{y}{2}} \\ \frac{dy}{dx} &= \frac{4x}{y} \\ m &= \frac{4x}{y}\end{aligned}$

another method
$\begin{aligned} x^2 - \frac{y^2}{4} &= 1 \\ \frac{y^2}{4} &= x^2 -1 \\ y &= \pm\sqrt{4(x^2-1)} \\ \frac{dy}{dx} &= \pm\frac{2x}{\sqrt{x^2-1}}\end{aligned}$

Homer · « **Reply #988 on:** December 10, 2012, 11:20:51 pm »

The options I am provided with are

a) any real number
b) any real number in the interval (-2,2)
c) any real number in the interval [-2,2]
d) any real number in the number R/(-2,2)
e) any real number in the interval R/[-2,2]

b^3 · « **Reply #989 on:** December 10, 2012, 11:28:32 pm »

I might have a slightly different interpretation of this, initally I went to use implicit differentiation too, then realised that they probably haven't come across it yet. But the way its worded, it may be asking for the values that m can be, which you can get from the second method, graph it and you can see that the gradient can be $R\backslash\left[-2,2\right]$ . To get this via another method, if you graph the original function, we can see it has asymptotes of $y=2x$ and $y=-2x$ . And when we look at the graph, looking at the right arm, the gradient is going to always be greater than 2 when it approaches $y=2x$ and always going to be less than (well more negative) than $y=-2x$ . For the left arm the gradient on the top part of the curve is under the asymptote and moving away from it, so that the gradient is always less than (more negative) than -2 and for the bottom half, always going to be greater than 2 (for the $y=2x$ asymtptote). SO that is $m\in R\backslash\left[-2,2\right]$ .

Anyways, thats just another way of looking at it, less mathematical though, but might help you understand it a little bit more.

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