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Sagot :
The equation of parabola is a [tex]\frac{-(x+ 3)^{2}}{64} + \frac{ (y-1)^{2}}{4} = 1[/tex] and the coordinates of center is (0,0) And asymptote is become 16.
According to the statement
we have to find the centers, foci, vertices and asymptotes of the hyperbola from the given equations.
So, For this purpose,
Hyperbola is a a plane curve generated by a point so moving that the difference of the distances from two fixed points is a constant.
So, The given equation is:
16y²-x² - 6x - 32y = 57
So, rearrange it then
-x² - 6x - 32y + 16y² = 57
-(x² + 6x) + (16y² - 32y ) = 57
-(x² + 6x) + 16(y² - 2y ) = 57
-(x² + 6x + 9) + 16(y² - 2y +1 ) = 57 +1(16) + 9(-1)
-(x+ 3)² + 16 (y-1)² = 64
-(x+ 3)² + 16 (y-1)² = 64
divide whole equation by 64 then
[tex]\frac{-(x+ 3)^{2}}{64} + \frac{16 (y-1)^{2}}{64} =\frac{64}{64}[/tex]
then equation become
[tex]\frac{-(x+ 3)^{2}}{64} + \frac{ (y-1)^{2}}{4} = 1[/tex]
This become the equation of hyperbola.
So,
here coordinates of center is (0,0)
And asymptote is become 16.
So, The equation of parabola is a [tex]\frac{-(x+ 3)^{2}}{64} + \frac{ (y-1)^{2}}{4} = 1[/tex] and the coordinates of center is (0,0) And asymptote is become 16.
Learn more about Hyperbola here
https://brainly.com/question/16454195
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