What is 4x^2+y^2+48x+128=0 ?

The solution to 4x^2+y^2+48x+128=0 is Ellipse with (h,k)=(-6,0),a=2,b=4

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4x^2+y^2+48x+128=0

4 x^{2} + y^{2} + 48 x + 128 = 0

(h, k) = (- 6, 0), a = 2, b = 4

Solution steps

4 x^{2} + y^{2} + 48 x + 128 = 0

Rewrite

4 x^{2} + y^{2} + 48 x + 128 = 0

in the form of the standard ellipse equation

\frac{( x - ( - 6 ) ) ^{2}}{2 ^{2}} + \frac{( y - 0 ) ^{2}}{4 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- 6, 0), a = 2, b = 4

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (- 6, 0), b = 4, a = 2

\frac{( x - 2 ) ^{2}}{9} + \frac{( y + 4 ) ^{2}}{16} = 1

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

64 xx - 256 x + 81 yy + 810 y = 2903

\frac{( x + 4 ) ^{2}}{9} + \frac{( y - 4 ) ^{2}}{16} = 1

(\frac{x - 1}{36})^{2} + (\frac{y + 2}{25})^{2} = 1

What is 4x^2+y^2+48x+128=0 ?
The solution to 4x^2+y^2+48x+128=0 is Ellipse with (h,k)=(-6,0),a=2,b=4