Ho Ho... No

[blacksanta62]

You're going to have to use the quadratic formula to get the intercepts. Gonna get a tad ugly and tedious, but do-able by hand (although you shouldn't be finding any x-intercepts ).

Okay, yeah the quadratic formula does get messy . How would I know where the points go on the graph if I get a number like: (0,2sqrt9) ? Would I have to assume it's a small number of would I be able to not have a scale on the graph? Or, in an exam situation, would VCAA ensure numbers that work out nicely? I have a lot of questions And why no x intercepts? Do you mean just for this graph or for every graphs?

[blacksanta62]

Yes, you can graph on the CAS!

Open a graph page > menu > graph entry (3) > equation (2) > ellipse (4)

Then just enter all your values in

btw that's for the TI-nspire

😲 That's the one I have... lol I never saw that equation thing, I'm going to have to play around with it alot!

[pi]

Okay, yeah the quadratic formula does get messy . How would I know where the points go on the graph if I get a number like: (0,2sqrt9) ? Would I have to assume it's a small number of would I be able to not have a scale on the graph? Or would VCAA ensure numbers that work out nicely? I have a lot of questions And why no x intercepts? Do you mean just for this graph or for every graphs?

Well it's a sketch, so it doesn't have to be exactly to scale haha! VCAA may or may not use nice numbers, hard to say.

There are no x-ints to find because this graph doesn't go through the x-axis (according to the quick sketch I did in my head anyway, could be wrong!)

[blacksanta62]

Ahh, so if they wanted it to scale they would provide a graph section (you know, a nice ruled up piece of paper, btw I like to be specfic ). A sketch is just freehand(?) and would be usually be done in textbook questions to warm you up?.

I should probably just practice with nice and not nice numbers anyway to be ready for thing. You're right about it not touching the x-axis though. Did the head sketches just come after a lot of questions and practice or were you always a visual person i.e. you see an equation and can just plot it in your head long before pencil touches paper

[pi]

I guess I'm a visual person (but still super average at maths compared to most on AN haha), but also I was taught how to sketch graphs very well in school which is why I can still do these spesh ones with no issues even though i haven't done any maths for about 4 years haha (other parts of the course though... no thanks LOL)

[Sine]

Along with that question: This is the equation for a graph- (x+3)^2/25 + (y-2)^2/16 = 1
The centre of the equation is (-3,2) but how does one find the min & max of the ellipse? Do I have to sketch it first? This is a multiple choice question so a CAS can be used but does the calculator have the ability to graph ellipses?

an ellipse (x-h)^2/a^2 + (y-k)^2/b^2 = 1
max y values
max (h,k+b) 
min  (h,k-b)

so for (x+3)^2/25 + (y-2)^2/16 = 1
max y values:
max occurs at : (-3,6)
min occurs at  : (-3,-2)

Sketch the graph of the following functions:
(x-2)^2/9 + (y+4)^2/4 = 1

I'm having trouble finding the x and y intercepts. The textbook says to make y = 0 & x = 0 (which I could normally do but I'm just lost with how the textbook got it's answer) but how would I do it with such a question? The question recommends no CAS be used. If some steps can be shown it will greatly appreciated. Thank you
 

(x-2)^2/9 + (y+4)^2/4 = 1
let x=0
(0-2)^2/9 + (y+4)^2/4 = 1
4/9 + (y+4)^2/4 = 1
(y+4)^2/4 = 5/9
(y+4)^2 = 20/9
y=+ or - 2sqrt{5}/3 -4

(x-2)^2/9 + (y+4)^2/4 = 1
let y=0
(x-2)^2/9 + (0+4)^2/4 = 1
(x-2)^2/9 + 16/4 = 1 
(x-2)^2/9 = -3
(x-2)^2= -27
no x intercepts (cant take square root of negative number)

[blacksanta62]

"other parts of the course though... no thanks"... lmao!
Something just occurred to me, if using the quadratic equation how would I turn it into a quadratic function? Multiply both sides by the denominators to get rid of them and then expand the brackets? so I would get: (x-2)^2 + (x+4)^2 = 324? and then 2x^2 + 4x + 20 = 324?

[blacksanta62]

Sine! Thanks for that but what about my random brain swirl above? Is it even possible?? I just spammed it with randomness

After reading you steps sine I understand what you did now to get the y and why there's no x Thank you all

*the first sentence sounded rude so I had to keep modifying to un-rude-it which sends me back to the forums page

[blacksanta62]

Just a tad bit confused on this part; (y+4)^2 = 20/9 ------> y=+ or - 2sqrt{5}/3 -4
Noob @ maths

[Sine]

"other parts of the course though... no thanks"... lmao!
Something just occurred to me, if using the quadratic equation how would I turn it into a quadratic function? Multiply both sides by the denominators to get rid of them and then expand the brackets? so I would get: (x-2)^2 + (x+4)^2 = 324? and then 2x^2 + 4x + 20 = 324?

what happened to the y in the ellipse? 

Just a tad bit confused on this part; (y+4)^2 = 20/9 ------> y=+ or - 2sqrt{5}/3 -4
Noob @ maths

(y+4)^2 = 20/9
take square root of both sides
y+4 = + or - sqrt{20/9}
simplify square roots
y+4 = + or - sqrt{20/9}
y+4 = + or - (sqrt{20}/sqrt{9})
sqrt 9 = 3 , sqrt 20 = sqrt{4} x sqrt{5} = 2 x sqrt{5}
y+4 = + or - 2sqrt{5}/3
 y=+ or - 2sqrt{5}/3 -4

[blacksanta62]

I don't even know where the y went... probably went to go join MCA, ba dum tss

Ahh because sqrt20 is a surd and 9 ---> 3, I normally write down more steps but you just went... Bam! Thanks sine.

P.s. Getting near that 100th post   

[blacksanta62]

Of course my 100th post will be regarding a specialist maths question
Sorry for all the coordinate geometry questions guys & girls: Give a pair of parametric equations which corresponds to the following Cartesian equation - (x+3)^2/25 + (y-1)^2/9 = 1

 

[Sine]

Of course my 100th post will be regarding a specialist maths question
Sorry for all the coordinate geometry questions guys & girls: Give a pair of parametric equations which corresponds to the following Cartesian equation - (x+3)^2/25 + (y-1)^2/9 = 1

Ellipse so sin^2(t) + cos^(t) = 1

sin^2(t) = (x+3)^2/25
sin(t) = x+3/5
x=5sin(t) - 3

cos^2(t) = (y-1)^2/9
cos(t) = y-1/3
y=3cos(t) + 1

Note there could be another pair of parametric equations by using cos for the x terms and sin for the y terms.

[blacksanta62]

This may be a 'simple' question but I was wondering if someone could help me
So I started the sketching hyperbolas section and felt great about it, so similar to circles and ellipses. These questions I found simple: x^2/4 - y^2/9 = 1 etc.

Need help with questions like these: (x-2)^2/25 - (y-3)^2/4 = 1
a^2 = 25 ---> a = 5
b^2 = 4 ----> b = 2
Centre: (2,3)
I though the asymptote  would be at y = -2/5x and y = 2/5x (going off the textbook examples but graphed on desmos and it wasn't. It also doesn't cross the x axis at -4 and 4 but at -7 and 7. 
Thank you

[byCrypt]

Need help with questions like these: (x-2)^2/25 - (y-3)^2/4 = 1
a^2 = 25 ---> a = 5
b^2 = 4 ----> b = 2
Centre: (2,3)
I though the asymptote  would be at y = -2/5x and y = 2/5x (going off the textbook examples but graphed on desmos and it wasn't. It also doesn't cross the x axis at -4 and 4 but at -7 and 7. 
Thank you

The general rule for asymptotes is: y=(b/a)(x-h)+k and y=(-b/a)(x-h)+k

The a and b values and the centre you have worked out are correct. In this case, h=2 and k=3

So you should end up with asymptotes y=(2/5)(x-2)+3 and y=(-2/5)(x-2)+3 (expanding these expressions should give you the correct asymptotes)