What is (x^2}{64}+\frac{(y-3)^2)/9 =1 ?

The solution to (x^2}{64}+\frac{(y-3)^2)/9 =1 is Ellipse with (h,k)=(0,3),a=8,b=3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x^2}{64}+\frac{(y-3)^2)/9 =1

\frac{x ^{2}}{64} + \frac{( y - 3 ) ^{2}}{9} = 1

(h, k) = (0, 3), a = 8, b = 3

Solution steps

\frac{x ^{2}}{64} + \frac{( y - 3 ) ^{2}}{9} = 1

Rewrite

\frac{x ^{2}}{64} + \frac{( y - 3 ) ^{2}}{9} = 1

in the form of the standard ellipse equation

\frac{( x - 0 ) ^{2}}{8 ^{2}} + \frac{( y - 3 ) ^{2}}{3 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (0, 3), a = 8, b = 3

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (0, 3), a = 8, b = 3

4 x^{2} + z^{2} = 16

\frac{( x - 5 ) ^{2}}{9} + \frac{( y - 2 ) ^{2}}{36} = 1

100 x^{2} + 64 y^{2} - 100 x - 1575 = 0

x^{2} + \frac{y ^{2}}{29} = 1

100 (x^{2} - x) + 16 (y^{2} - 4 y) = 147

What is (x^2}{64}+\frac{(y-3)^2)/9 =1 ?
The solution to (x^2}{64}+\frac{(y-3)^2)/9 =1 is Ellipse with (h,k)=(0,3),a=8,b=3