What is the vertices ((x-3)^2)/(16)-((y+4)^2)/(25)=1 ?

The vertices ((x-3)^2)/(16)-((y+4)^2)/(25)=1 is (7,-4),(-1,-4)

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vertices ((x-3)^2)/(16)-((y+4)^2)/(25)=1

vertices

\frac{( x - 3 ) ^{2}}{16} - \frac{( y + 4 ) ^{2}}{25} = 1

(7, - 4), (- 1, - 4)

Solution steps

(h + a, k), (h - a, k)

Calculate Hyperbola properties

\frac{( x - 3 ) ^{2}}{16} - \frac{( y + 4 ) ^{2}}{25} = 1 :

Right-left Hyperbola with

(h, k) = (3, - 4), a = 4, b = 5

(3 + 4, - 4), (3 - 4, - 4)

Refine

(7, - 4), (- 1, - 4)

5 x^{2} - 5 y^{2} + 20 x - 60 y = 185

9 x^{2} - 4 y^{2} + 54 x - 8 y + 78 = 0

x^{2} - y^{2} < 1

x^{2} - y^{2} = 9

4 y^{2} - x^{2} + 40 y - 4 x + 60 = 0

What is the vertices ((x-3)^2)/(16)-((y+4)^2)/(25)=1 ?
The vertices ((x-3)^2)/(16)-((y+4)^2)/(25)=1 is (7,-4),(-1,-4)