(x^2/81)-(y^2/4)=1 Find the Foci of the parabola
Are you sure this is a parabola? Both variables are squared.
Oops, I meant hyperbola!
Hyperbola, exactly :P \[\frac{x^2}{81}-\frac{y^2}{4}=1\] => a=√81 = 9 and b= √4 = 2 For a hyperbola, \[a^2 + b^2 = c^2\] And it's focus is given by : (±c,0)
What do you get for vertices? @Aditi_Singh
Erm .. The vertices would be : (±a,0)
So, that is (-9,0) and (9,0) vertices. And, (√85, 0) and (-√85,0) for foci.
Yep!
@Aditi_Singh Would you give your opinion on another hyperbola question? Three of us are not in agreement about the foci of this hyperbola. I'll post the thread here in the event you are up for the challenge. :)
Haha :D please :) :P
I found the question, multiple choice, too. Find the coordinates of the foci of the hyperbola y^2/9-x^2/16=1. A) (4, 0) and (-4, 0) B) (5, 0) and (-5, 0) C) (0, square root 7) and (0, -square root 7) D) (0, 5) and (0, -5)
I would go with 'D' - (0,±5)
Alrighty, now for the big "reveal."
because, here the foci is on y axis ..
*Fingers crossed*
We're together.
Hi5! xD
@sakigirl Here's a conic section practice test with answers. http://www.education.com/study-help/article/pre-calculus-help-pre-calculus-12-conic/
@Directrix Thanks!
@Aditi_Singh I tried three times to tell the Asker but she ignored me and went with the others. That two of the options had 5s in the them was a BIG clue.
@sakigirl You could post those questions one by one here, tell in the post that you are practicing for a test tomorrow, and practice that way.
Awh! I just went through the whole question .. And .. I .... I .. I'm speechless :P
My gosh. Out of desperation, I posted the graph. But, no, I was ignored.
Hahah :P Ignore them - as simple as that :P
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