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Mathematics 10 Online
OpenStudy (anonymous):

Could someone help me please?

OpenStudy (anonymous):

OpenStudy (anonymous):

One second.

OpenStudy (anonymous):

It's not A.

OpenStudy (anonymous):

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!$$\pm\frac34(x+2)=y-3\\y=\pm\frac34(x+2)+3$$

OpenStudy (anonymous):

BTW -- this tells us \((-2,3)\) is our center :-) those lines are our asymptotes.

OpenStudy (anonymous):

I still don't understand the vertices though.

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