What is the vertices-9x^2+16y^2-36x-160y+220=0 ?

The vertices-9x^2+16y^2-36x-160y+220=0 is (-2,8),(-2,2)

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vertices-9x^2+16y^2-36x-160y+220=0

vertices

- 9 x^{2} + 16 y^{2} - 36 x - 160 y + 220 = 0

(- 2, 8), (- 2, 2)

Solution steps

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

Calculate Hyperbola properties

- 9 x^{2} + 16 y^{2} - 36 x - 160 y + 220 = 0 :

Up-down Hyperbola with

(h, k) = (- 2, 5), a = 3, b = 4

(- 2, 5 + 3), (- 2, 5 - 3)

Refine

(- 2, 8), (- 2, 2)

\frac{y ^{2}}{4} - \frac{x ^{2}}{21} = 1

\frac{x ^{2}}{49} - \frac{y ^{2}}{81} = 1

\frac{x ^{2}}{49} - \frac{y ^{2}}{36} = 1

\frac{y ^{2}}{25} - \frac{x ^{2}}{64} = 1

9 y^{2} - 25 x^{2} = 225

What is the vertices-9x^2+16y^2-36x-160y+220=0 ?
The vertices-9x^2+16y^2-36x-160y+220=0 is (-2,8),(-2,2)