What is 27x^2+6y^2-5=0 ?

The solution to 27x^2+6y^2-5=0 is Ellipse with (h,k)=(0,0),a=(sqrt(5))/(3sqrt(3)),b=sqrt(5/6)

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

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

(h, k) = (0, 0), a = \frac{5}{3 3}, b = \frac{5}{6}

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

\frac{( x - 0 ) ^{2}}{( \frac{5}{3 3} ) ^{2}} + \frac{( y - 0 ) ^{2}}{( \frac{5}{6} ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (0, 0), a = \frac{5}{3 3}, b = \frac{5}{6}

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (0, 0), b = \frac{5}{6}, a = \frac{5}{3 3}

- 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

\frac{x ^{2}}{9} + \frac{y ^{2}}{4} = 1 \cdot \frac{1}{9} \cdot \frac{1}{4}

What is 27x^2+6y^2-5=0 ?
The solution to 27x^2+6y^2-5=0 is Ellipse with (h,k)=(0,0),a=(sqrt(5))/(3sqrt(3)),b=sqrt(5/6)