What is the vertices 4x^2-y^2-16x-6y-9=0 ?

The vertices 4x^2-y^2-16x-6y-9=0 is (4,-3),(0,-3)

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vertices 4x^2-y^2-16x-6y-9=0

vertices

4 x^{2} - y^{2} - 16 x - 6 y - 9 = 0

(4, - 3), (0, - 3)

Solution steps

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

Calculate Hyperbola properties

4 x^{2} - y^{2} - 16 x - 6 y - 9 = 0 :

Right-left Hyperbola with

(h, k) = (2, - 3), a = 2, b = 4

(2 + 2, - 3), (2 - 2, - 3)

Refine

(4, - 3), (0, - 3)

4 x^{2} - 9 y^{2} = 144

f (x) f (x) = 3 (x + 6) (x + 3)

4 x^{2} - 9 y^{2} - 16 x + 18 y - 7 = 0

\frac{x ^{2}}{7} - \frac{y ^{2}}{6} = 1

9 x^{2} - 16 y^{2} + 54 x - 16 y = - 113

What is the vertices 4x^2-y^2-16x-6y-9=0 ?
The vertices 4x^2-y^2-16x-6y-9=0 is (4,-3),(0,-3)