Hey there.
May is always a busy month for most VCE students, so hang in there - it's only temporary
Some advice: Do remember that for this year, you have two subjects that will contribute to your ATAR, and so those should be your priority. Note that I'm
NOT saying you should neglect your year 11 studies. I'm just trying to remind you of what is important, so do make sure you are feeling prepared and confident for Methods and Bio first.
How do you find the range of an ellipse?
Consider the following relation: \[\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1,\quad a,b>0.\] This is an ellipse that
> is centered at \((h,\,k)\),
> has a horizontal semi-axis length of \(a\), and
> has a vertical semi-axis length of \(b\).
From the following image, it's quite easy to find the domain and range.
How do you prove that x^2 +y^2+a^2+2axy= 1 touches the x axis (or y axis- can't remember which axis the question referred to)?
Unfortunately, I'm not sure you remembered the relation correctly, since for \[C:\ \ x^2+y^2+a^2+2axy=1,\] the graph either crosses the coordinate axes, or doesn't touch the coordinate axes at all (depending on the value of \(a\)). That is, for all \(a\in\mathbb{R}\), the graph never touches either axis without crossing.
There are a few ways you can go about proving that a relation only touches either coordinate axis. For example, for conic sections such as circles and ellipses, you could find the 'domain' and range. Take the graph of \((x-2)^2+y^2=4\), which is a circle of radius 2 centered at \((2,\,0)\). So, it has domain \([0,\,4]\) and range \([-2,\, 2]\), and therefore touches (but does not cross) the \(y\)-axis.
How do you find tangents to circles?
There are a couple ways to go about this. Let \(C\) denote the centre of the circle and let \(P\) be the point on the circle where the tangent touches. The simplest way to find the equation of the tangent is to realise that the tangent will be
perpendicular to the line segement \(CP\).
For example, let's find the equation of the tangent to the graph of \((x-2)^2+y^2=4\) at the point \(P(1,\,\sqrt{3})\). The gradient of the line joining \(P\) and the centre of the circle \(C(2,\,0)\) is \[m_{CP}=\frac{\sqrt{3}-0}{1-2}=-\sqrt3,\] and so the gradient of the tangent to the graph at \(P\) is \(\dfrac{1}{\sqrt{3}}\). Hence, the equation of the tangent is \[y-\sqrt{3}=\frac{1}{\sqrt{3}}(x-1)\\
\implies y=\frac{1}{\sqrt{3}}x+\frac{2\sqrt3}{3}.\]
Could someone explain how to solve modulus equations if they are literal (I don't remember much more about the question than that unfortunately)
Recall that \[|x|=\begin{cases}x, & x\geq 0\\
-x, & x<0\end{cases}.\] You can essentially break up the problem into cases. Suppose we want to solve \[|ax-b|=c,\quad a>0,\ \ b\in\mathbb{R},\ \ c\geq 0.\] Then, using our definition, we have \[ax-b=c\ \text{ if }x\geq\frac{b}{a}\quad\text{and}\quad b-ax=c\ \text{ if }x<\frac{b}{a}\\
\implies x=\frac{b+c}{a}\ \text{ or }\ x=\frac{b-c}{a}.\] Alternatively, you can use \(|x|=\sqrt{x^2}\) if it makes it easier (note that sometimes, it won't).
Is there some "secret" to solving complex bearings problems?
Not really. You get more efficient at them with practice. It's not like there's a trick so that you can see solutions immediately unfortunately.
How do you do parametric equations?
To be honest, I'm not sure what you mean by
"do" parametric equations. There are lots of things you can do with them. What specifically are you looking for?
Hope this has been helpful
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