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

The vertices (x^2)/(4^2)+(y^2)/(3^2)=1 is (4,0),(-4,0)

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

vertices

\frac{x ^{2}}{4 ^{2}} + \frac{y ^{2}}{3 ^{2}} = 1

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

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{4 ^{2}} + \frac{y ^{2}}{3 ^{2}} = 1 :

Ellipse with center

(h, k) = (0, 0), a = 4, b = 3

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

Refine

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

25 x^{2} + 9 y^{2} - 200 x - 36 y + 211 = 0

x^{2} + y^{2} + 6 x - 4 y - 3 = 0

\frac{x ^{2}}{100} + \frac{y ^{2}}{49} = 1

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

x^{2} + 2 y^{2} - 1 = 0

What is the vertices (x^2)/(4^2)+(y^2)/(3^2)=1 ?
The vertices (x^2)/(4^2)+(y^2)/(3^2)=1 is (4,0),(-4,0)