What is the vertices ((x-2)^2)/9+((y-3)^2)/(49)=1 ?

The vertices ((x-2)^2)/9+((y-3)^2)/(49)=1 is (2,10),(2,-4)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices ((x-2)^2)/9+((y-3)^2)/(49)=1

vertices

\frac{( x - 2 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{49} = 1

(2, 10), (2, - 4)

Solution steps

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

Calculate ellipse properties

\frac{( x - 2 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{49} = 1 :

Ellipse with center

(h, k) = (2, 3), b = 7, a = 3

(2, 3 + 7), (2, 3 - 7)

Refine

(2, 10), (2, - 4)

9 x^{2} + 81 y^{2} - 729 = 0

16 x^{2} + 4 y^{2} = 64

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

\frac{( x - 3 ) ^{2}}{4} + \frac{( y - 4 ) ^{2}}{25} = 1

16 x^{2} - 64 x + 9 y^{2} - 54 y = - 1

What is the vertices ((x-2)^2)/9+((y-3)^2)/(49)=1 ?
The vertices ((x-2)^2)/9+((y-3)^2)/(49)=1 is (2,10),(2,-4)