What is 16x^2-32x+9y^2=128 ?

The solution to 16x^2-32x+9y^2=128 is Ellipse with (h,k)=(1,0),a=3,b=4

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16x^2-32x+9y^2=128

16 x^{2} - 32 x + 9 y^{2} = 128

(h, k) = (1, 0), a = 3, b = 4

Solution steps

16 x^{2} - 32 x + 9 y^{2} = 128

Rewrite

16 x^{2} - 32 x + 9 y^{2} = 128

in the form of the standard ellipse equation

\frac{( x - 1 ) ^{2}}{3 ^{2}} + \frac{( y - 0 ) ^{2}}{4 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (1, 0), a = 3, b = 4

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (1, 0), b = 4, a = 3

169 x^{2} + 144 y^{2} - 338 x - 864 y - 22871 = 0

x^{2} + y^{2} < 2 y

25 x^{2} + 50 y^{2} = 25

\frac{4 x ^{2}}{25} + \frac{9 y ^{2}}{25} = 1

X^{2} + 8 Y^{2} - 5 = 0

What is 16x^2-32x+9y^2=128 ?
The solution to 16x^2-32x+9y^2=128 is Ellipse with (h,k)=(1,0),a=3,b=4