9>x^2+9y^2

Solution

9 > x^{2} + 9 y^{2}

(h, k) = (0, 0), a = 3, b = 1

Solution steps

9 > x^{2} + 9 y^{2}

Rewrite

9 > x^{2} + 9 y^{2}

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (0, 0), a = 3, b = 1

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (0, 0), a = 3, b = 1

27 x^{2} + 6 y^{2} - 5 = 0

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

\frac{u + v}{u ^{2} + v ^{2}} + \frac{- v + u}{u ^{2} + v ^{2}} = 1

12 (x + 4)^{2} + 3 (y - 1)^{2} = 48

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

What is 9>x^2+9y^2 ?
The solution to 9>x^2+9y^2 is Ellipse with (h,k)=(0,0),a=3,b=1