What is the vertices 4x^2+16y^2-16x-48=0 ?

The vertices 4x^2+16y^2-16x-48=0 is (6,0),(-2,0)

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vertices 4x^2+16y^2-16x-48=0

vertices

4 x^{2} + 16 y^{2} - 16 x - 48 = 0

(6, 0), (- 2, 0)

Solution steps

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

Calculate ellipse properties

4 x^{2} + 16 y^{2} - 16 x - 48 = 0 :

Ellipse with center

(h, k) = (2, 0), a = 4, b = 2

(2 + 4, 0), (2 - 4, 0)

Refine

(6, 0), (- 2, 0)

25 (x + 2)^{2} + 4 (y - 3)^{2} = 100

4 x^{2} + y^{2} + 6 y + 5 = 0

\frac{( x + 2 ) ^{2}}{16} + \frac{y ^{2}}{49} = 1

(\frac{x}{7})^{2} + (\frac{y}{5})^{2} = 1

36 x^{2} + 25 y^{2} - 50 y - 875 = 0

What is the vertices 4x^2+16y^2-16x-48=0 ?
The vertices 4x^2+16y^2-16x-48=0 is (6,0),(-2,0)