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Popular >

vertices 3x^2+4y^2=48

Solution

vertices

3 x^{2} + 4 y^{2} = 48

Solution

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

Solution steps

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

Calculate ellipse properties

3 x^{2} + 4 y^{2} = 48 :

Ellipse with center

(h, k) = (0, 0), a = 4, b = 23

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

Refine

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

Popular Examples

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2 x^{2} + y^{2} - 98 = 0

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3 x^{2} - 6 x + y^{2} + 4 y - 11 = 0

vertices ((x+2)^2)/9+((y-3)^2)/5 =1 vertices

\frac{( x + 2 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{5} = 1

vertices (x^2)/(144)+(y^2)/(169)=1 vertices

\frac{x ^{2}}{144} + \frac{y ^{2}}{169} = 1

vertices 9x^2+y^2-18x-6y+9=0 vertices

9 x^{2} + y^{2} - 18 x - 6 y + 9 = 0

Frequently Asked Questions (FAQ)

What is the vertices 3x^2+4y^2=48 ?
The vertices 3x^2+4y^2=48 is (4,0),(-4,0)