Sketching Standard Form Hyperbolas (Conics)

Sorry for the lateness of this posting. I had a blonde moment. The video was showing up as text but I guess when I publish it, it shows up as normal. Anyway....



Today we learned how to graph hyperbolas.....YAAAAAAAY!, Lots of fun.......but nowhere as much fun as trig, because it's awesome.


The first slide is just showing you the equation of a hyperbola. A great way to look at it is that either the x coordinate or y coordinate is negative. Also, the coordinate that is positive is the way that the hyperbola opens. So, if you have +x and -y, the hyperbola is going to open left-right because your x coordinate is positive. Furthermore, your 'a' value always goes with your positive value (x or y); However, unlike circles, your 'a' value (transverse axis=2a) is NOT always bigger than your 'b' value (conjugate axis=2b).

Moving on, the next slide asks you to sketch the hyperbola.
The first thing that I do is find the centre of the hyperbola. This is the same as finding the centre for a line. It's just the opposite sign of what's inside the brackets. In this case it is (-2,1).
Next, I find the values for 'a' and 'b'. For this, you just have to square root the denominator to get your values. Now, while keeping in mind which way the hyperbola opens, you can trace your transverse and conjugate axes. I recommend drawing a box for your axes, it really helps. From there you can find your vertecies which are the endpoints for transverse axis. The vertecies are your starting points for your hyperbola.
Now, we have to find the equation of the asymptotes to make our sketch complete and where to draw our hyperbola. This is just finding the equation of a line: y=mx+b, going back to grade 10 if you remembered which I didn't. So, you know your x and y values, they're just the values of the centre (x=-2,y=1). You know the m (slope), it's (by the way that I look at it) the amount that you move from the centre to the corner of the box. In this case, you're moving two up and three over making the slope 2/3 and -2/3. The part that you don't know is b (different from b at the start), but since you know everything else, it is just a matter of putting everything in and solving. After you find your b values, you can draw your asymptotes equations (with dashed lines) and then draw your hyperbola, remembering that your hyperbola follows very close to your asymptotes.

The next slide is just another example.

Last, you are given the graph and it is just a matter of finding the equation.
What I do first is find out whether your x or y is negative by looking at which way the hyperbola opens.
Next, I find the centre of the graph; In this case it is (-4,1),making it(x+4)-(y-1).
Now, you have to find the value of 'a' and 'b'. Without aid of the box or it telling you what your endpoints of the conjugate axes are, you can't find your 'b' value. Since you know the vertecies of the graph, you can determine the value of 'a', but you have to put it in terms of a^2 as opposed to 2a. In this case, you're given a box so you can find the 'b' vale.
Last, you have to find the equation of the asymptotes which is the same as before since you're given the box.

Hoped this helped you out if you were having troubles with this.

Conics- major / minor axis ( from formulas) - Transverse / conjugate axis

For ellipses, you have two axis, a major and a minor . the major axis is always the longer of the two ( see video, starts on ellipses).





For Hyperbolas, you also have two axis, a transverse axis and a conjugate axis. ( see video, after ellipses)





Transverse axis


-connects verticies


-not always longer than conjugate axis


- length is alwas 2a units (a+a)


- there for length of conjugate is always 2b units (b+b)

More Conics

Today in class we talked more about conics. Last class we found out how to change from general form to standard form. This class it was the complete opposite, we learned how to go from standard form back to general form. The reason we convert from general form to standard form is to be able to graph. I didn't quite get the concept at the start of the conics lesson but I did some accelerated math and I am starting to understand it. Also the test has been changed to next Thursday.