What is the vertices 49x^2+25y^2-196x+250y-404=0 ?

The vertices 49x^2+25y^2-196x+250y-404=0 is (2,2),(2,-12)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices 49x^2+25y^2-196x+250y-404=0

vertices

49 x^{2} + 25 y^{2} - 196 x + 250 y - 404 = 0

(2, 2), (2, - 12)

Solution steps

(h, k + b), (h, k - b)

Calculate ellipse properties

49 x^{2} + 25 y^{2} - 196 x + 250 y - 404 = 0 :

Ellipse with center

(h, k) = (2, - 5), b = 7, a = 5

(2, - 5 + 7), (2, - 5 - 7)

Refine

(2, 2), (2, - 12)

\frac{( x - 3 ) ^{2}}{49} + \frac{( y + 5 ) ^{2}}{25} = 1

\frac{x ^{2}}{64} + \frac{y ^{2}}{100} = 1

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

4 x^{2} + 9 y^{2} - 16 x + 18 y - 11 = 0

\frac{x ^{2}}{16} + \frac{y ^{2}}{12} = 1

What is the vertices 49x^2+25y^2-196x+250y-404=0 ?
The vertices 49x^2+25y^2-196x+250y-404=0 is (2,2),(2,-12)