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So I have this equation: 36x^2-121y^2=49 Which I rearranged to 36x^2/49-121y^2/49=1 So I want to go from here and find the eccentricity,foci,asymptotes etc...
But I'm getting myself confused. Since my equation is of the form x^2/a^2-y^2/b^2=1
Both a and b would be 7^2... which makes my eccentricity e=sqrt(1+49/49) but this is at odds with the software I'm using to practice this.
Which part am I getting myself tied up with? Thank you!

Respuesta :

mrinconj
Well, as the equation form is x^2/a^2-y^2/b^2=1 you have to have in each numerator only x^2 and y^2 but you have 36x^2 and 121y^2, so you have to work more for getting the equation form and i have it for you:
First divide your equation by 36:    36x^2/36-121y^2/36=49/36
you get  x^2-121y^2/36=49/36
Then divide that by 121:   x^2/121-y^2/36=49/4356
 Now divide this by 49/4356:    x^2/(49/36)-y^2/(49/121)=1
Finally sqrt and square for denominators and we will get the equation form:
x^2/(7/6)^2-y^2/(7/11)^2=1
From here you have a=7/6 and b=7/11
Now excentricity is e=(sqrt(a^2+b^2))/a "you have a mistake in this formula"
e=(sqrt((7/6)^2+(7/11)^2)/(7/6) = 1.139 ≈ 1.14 =57/70

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