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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: M-D on April 18, 2013, 02:30:06 pm

Title: sketch this hyperbola
Post by: M-D on April 18, 2013, 02:30:06 pm

i've got the following question:

-x^2-2x+4y^2-8y-1=0

i completed the square as follows:

-x^2-2x+4y^2-8y-1=0

-x^2-2x+4y^2-8y=1

-(x^2+2x)+4(y^2-2y)=1

-[(x^2+2x+1)-1]+4[(y^2-2y+1)-1]=1

-(x^2+2x+1)+1+4(y^2-2y+1)-4=1

-(x+1)^2+4(y-1)^2=4

i don't know what to next. can someone please help

Title: Re: sketch this hyperbola
Post by: Alwin on April 18, 2013, 03:18:40 pm

Step 1: CHILL  :P

Step 2: Put it in 'standard' form, -((x+1)^2)/4+(y-1)^2=1

Step 3: Find the centre of the hyperbola. You look at the translations, in you case it is left 1, up 1

Step 4: Work out the gradient of the asymptote, so rise over run. Your 'rise' is the sqrt value beneath the y-term, so 1. Your 'run' value is the sqrt value beneath the x-term, so 2. hence gradient is 1/2. From here you can find the equation of the asymptotes

Step 5: Check if it is up and down or left and right hyperbola. since the minus sign is on the x term, yours will be an up-and-down hyperbola.

Step 6: The vertices of up-and-down hyperbola is the centre y-value +- the sqrt value under the y term.

Hopefully that should be enough to get you well under way. The generic formula is probably in your textbook somewhere. A picture of the graph is: http://www4a.wolframalpha.com/Calculate/MSP/MSP4261aa8c7b9hb4g6401000050h6c7ffg8bc80a4?MSPStoreType=image/gif&s=25&w=200.&h=202.&cdf=RangeControl

Btw, have you just started this topic? we covered this over the summer holidays agess ago...

Title: Re: sketch this hyperbola
Post by: M-D on April 18, 2013, 05:43:52 pm

thanks a lot. No we have already covered the whole topic. I'm just doing a few practice questions.