What is the length of the major axis of the conic section shown below? (x-3)^2/49+ (y+6)^2/100=1 A. 20 B. 14 C. 10 D. 7
you know what this looks like?
I do not. I don't really remember this section. Conic sections went by to fast.
yeah they always stick em at the end when there is not enough time
in any case you don't really need to know that much for this question general form is \[\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1\] in your case \(a^2=49, a=7,b^2=100,b=10\)
since \(b=10\) is the bigger of the two the major axis has length \(20\) i.e. \(2\times 10\)
|dw:1407898076909:dw|
O.O is that the answer?
like that because the larger number is under the \(y\) term
yes
why would i lie?
xD Yhu don't really have a reason... I'm just wondering what the heck happened. ;o;
not sure what you mean in your case \(b=10\) and the major axis has length double that
Ok... I'l start with... Where did yhu get the 49 and 100 from? Nevermind... the denominators... why did yhu cut 'em in half? It said squared.
lets go slow
general form is \[\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1\] with center \((h,k)\) you have \[\frac{(x-3)^2}{49}+\frac{(y+6)^2}{100}=1\]
that makes the center \((3,-6)\) although you were not asked for the center
it also make \(a^2=49\) so you know \(a=7\) and similarly \(b^2=100\) making \(b=10\)
because the larger number is under the \(y\) term it looks like the one on the right, not the one on the left |dw:1407898559274:dw|
you weren't asked that either, but no matter you were asked for the length of the major axis the fact that \(b=10\) tells you that the distance from the center to the vertices is 10 units, making the length of the axis 20|dw:1407898660084:dw|
Oh my god. >.< I remember this now. Yhu helped me with the same thing a while ago... but I didn't have the graph so I got confused... Yhu like... A genius. And I deem yhu my favorite mod. Thank yhu for the help. ^u^ Again.
your welcome glad i am your fave mod
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