SOLUTION: Find the standard equation,foci,asymptotes and vertices of 25x^2-39y^2+150x+390y=-225

Question 1043331: Find the standard equation,foci,asymptotes and vertices of 25x^2-39y^2+150x+390y=-225
Answer by KMST(5328) (Show Source): You can put this solution on YOUR website!

The standard equation above tells us that
the center is at (-3,5) ;
the major (or transverse) axis is ;
the focal distance is ;
the asymptotes have slopes such that , and
the y-coordinates of the vertices can be found from
.
The asymptotes pass through center ,
and have slopes such as
--> ,
the equations of the asymptotes (in point-slope form) are
and .
Those equations can also be written as
and
in slope-intercept form.
Since the foci are on the major axis,
at a distance from the center,
their y- coordinates are , or .
So, one focus is at ,
and the tother focus is at
As for the vertices, also on the major axis,
-->-->---> .
So the vertices are at and .
The hyperbola with major axis and foci looks like this:

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