What is x^2+9y^2<= 16 ?

The solution to x^2+9y^2<= 16 is Ellipse with (h,k)=(0,0),a=4,b= 4/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^2+9y^2<= 16

x^{2} + 9 y^{2} \leq 16

(h, k) = (0, 0), a = 4, b = \frac{4}{3}

Solution steps

x^{2} + 9 y^{2} \leq 16

Rewrite

x^{2} + 9 y^{2} \leq 16

in the form of the standard ellipse equation

\frac{( x - 0 ) ^{2}}{4 ^{2}} + \frac{( y - 0 ) ^{2}}{( \frac{4}{3} ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (0, 0), a = 4, b = \frac{4}{3}

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (0, 0), a = 4, b = \frac{4}{3}

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

- (x - 2)^{2} - y^{2} = 4

\frac{x ^{2}}{5.6 6 ^{2}} + \frac{y ^{2}}{3.5 9 ^{2}} = 1

5 x^{2} + 2 y^{2} = 1

5 x^{2} + 2 y^{2} = 2

What is x^2+9y^2<= 16 ?
The solution to x^2+9y^2<= 16 is Ellipse with (h,k)=(0,0),a=4,b= 4/3