What is y^2+4x^2=128 ?

The solution to y^2+4x^2=128 is Ellipse with (h,k)=(0,0),a=4sqrt(2),b=8sqrt(2)

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y^2+4x^2=128

y^{2} + 4 x^{2} = 128

(h, k) = (0, 0), a = 42, b = 82

Solution steps

y^{2} + 4 x^{2} = 128

Rewrite

y^{2} + 4 x^{2} = 128

in the form of the standard ellipse equation

\frac{( x - 0 ) ^{2}}{( 4 2 ) ^{2}} + \frac{( y - 0 ) ^{2}}{( 8 2 ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (0, 0), a = 42, b = 82

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (0, 0), b = 82, a = 42

2 x^{2} + 3 y^{2} + 10 x + 9 y - 19 = 0

2 (x - 3)^{2} + 3 (y - 2)^{2} = 6

(\frac{x ^{2}}{15}) + ((\frac{( y - 8 )}{49})^{2}) = 1

100 - 25 x^{2} - 4 y^{2} \geq 0

(\frac{x - 3}{2})^{2} + (y - 7)^{2} = 5^{2}

What is y^2+4x^2=128 ?
The solution to y^2+4x^2=128 is Ellipse with (h,k)=(0,0),a=4sqrt(2),b=8sqrt(2)