What is 16x^2-32x+7y^2-33-42y=0 ?

The solution to 16x^2-32x+7y^2-33-42y=0 is Ellipse with (h,k)=(1,3),a=sqrt(7),b=4

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16x^2-32x+7y^2-33-42y=0

16 x^{2} - 32 x + 7 y^{2} - 33 - 42 y = 0

(h, k) = (1, 3), a = 7, b = 4

Solution steps

16 x^{2} - 32 x + 7 y^{2} - 33 - 42 y = 0

Rewrite

16 x^{2} - 32 x + 7 y^{2} - 33 - 42 y = 0

in the form of the standard ellipse equation

\frac{( x - 1 ) ^{2}}{( 7 ) ^{2}} + \frac{( y - 3 ) ^{2}}{4 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (1, 3), a = 7, b = 4

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (1, 3), b = 4, a = 7

\frac{( x - 1 ) ^{2}}{5} + \frac{( y - 2 ) ^{2}}{8} = 1

11 x^{2} + 2 y^{2} - 22 = 0

\frac{( y + 3 ) ^{2}}{81} + \frac{( x - 7 ) ^{2}}{64} = 1

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

9 x^{2} + 4 y^{2} + 36 x - 8 y + 4 = 0

What is 16x^2-32x+7y^2-33-42y=0 ?
The solution to 16x^2-32x+7y^2-33-42y=0 is Ellipse with (h,k)=(1,3),a=sqrt(7),b=4