Ask
your own question, for FREE!
Mathematics
10 Online
Could someone help me please?
Still Need Help?
Join the QuestionCove community and study together with friends!
One second.
It's not A.
What that equation basically tells us is that a certain "skewed" distance metric will equal \(1\) between every point on the ellipse and our point. For our center, we expect this distance to be \(0\) so we wish to solve:$$(x+2)^2/16-(y-3)^2/9=0$$and$$(x+2)^2/16=(y-3)^2/9\\9(x+2)^2=16(y-3)^2\\\pm3(x+2)=\pm4(y-3)$$ Observe the trivial solution \(x=-2,y=3\). Also observe that from above we can derive the equations of our asymptotes!$$\pm\frac34(x+2)=y-3\\y=\pm\frac34(x+2)+3$$
BTW -- this tells us \((-2,3)\) is our center :-) those lines are our asymptotes.
Still Need Help?
Join the QuestionCove community and study together with friends!
I still don't understand the vertices though.
Can't find your answer?
Make a FREE account and ask your own questions, OR help others and earn volunteer hours!
Join our real-time social learning platform and learn together with your friends!
Join our real-time social learning platform and learn together with your friends!
Latest Questions
VERNEISH:
Two objects are moving along separate linear paths where each path is described by position, d, and time, t.
Maci:
what are the best stock apps for people who want to start investing?
Twaylor:
genuienly how do i get better self control i keep watcing youtube and theres this
IzzyCookie:
|dw:1792074056138:dw| Uhm, I don't know where to ask this, but I'm in college welding at my community college, and I need help learning vertical welding for
thehearken:
not sure if this is considered spam, im sorry if it is, but does anyone have tips
1 hour ago
0 Replies
0 Medals
3 days ago
3 Replies
2 Medals
6 days ago
3 Replies
0 Medals
1 day ago
9 Replies
1 Medal
1 week ago
2 Replies
1 Medal