What is 8x^2+4y^2=21 ?

The solution to 8x^2+4y^2=21 is Ellipse with (h,k)=(0,0),a=(sqrt(21)}{2sqrt(2)},b=\frac{sqrt(21))/2

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8x^2+4y^2=21

8 x^{2} + 4 y^{2} = 21

(h, k) = (0, 0), a = \frac{21}{2 2}, b = \frac{21}{2}

Solution steps

8 x^{2} + 4 y^{2} = 21

Rewrite

8 x^{2} + 4 y^{2} = 21

in the form of the standard ellipse equation

\frac{( x - 0 ) ^{2}}{( \frac{21}{2 2} ) ^{2}} + \frac{( y - 0 ) ^{2}}{( \frac{21}{2} ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (0, 0), a = \frac{21}{2 2}, b = \frac{21}{2}

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (0, 0), b = \frac{21}{2}, a = \frac{21}{2 2}

\frac{( x - 1 ) ^{2}}{9} + \frac{( y + 2 ) ^{2}}{16} = 1

144 x^{2} + 169 y^{2} - 288 x + 676 y - 33516 = 0

\frac{( ( x + 2 ) ^{2} )}{4} + \frac{( ( y - 4 ) ^{2} )}{16} = 1

25 x^{2} + 50 x + 4 y^{2} - 16 y = 59

\frac{( x - 1 ) ^{2}}{4} + \frac{( y - 2 ) ^{2}}{16} = 1

What is 8x^2+4y^2=21 ?
The solution to 8x^2+4y^2=21 is Ellipse with (h,k)=(0,0),a=(sqrt(21)}{2sqrt(2)},b=\frac{sqrt(21))/2