What is 17x^2+16y^2-102x-32y-103=0 ?

The solution to 17x^2+16y^2-102x-32y-103=0 is Ellipse with (h,k)=(3,1),a=4,b=sqrt(17)

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17x^2+16y^2-102x-32y-103=0

17 x^{2} + 16 y^{2} - 102 x - 32 y - 103 = 0

(h, k) = (3, 1), a = 4, b = 17

Solution steps

17 x^{2} + 16 y^{2} - 102 x - 32 y - 103 = 0

Rewrite

17 x^{2} + 16 y^{2} - 102 x - 32 y - 103 = 0

in the form of the standard ellipse equation

\frac{( x - 3 ) ^{2}}{4 ^{2}} + \frac{( y - 1 ) ^{2}}{( 17 ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (3, 1), a = 4, b = 17

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (3, 1), b = 17, a = 4

x^{2} - 4 x + 100 y^{2} + 200 y + 4 = 0

2 y - xx - yy > 0

2 x^{2} + 4 y^{2} + 8 x + 9 y - 3 = 0

25 x^{2} + 9 y^{2} - 100 x - 54 y - 44 = 0

\frac{( x - 1 ) ^{2}}{9} + \frac{( y ^{2} )}{5} = 100

What is 17x^2+16y^2-102x-32y-103=0 ?
The solution to 17x^2+16y^2-102x-32y-103=0 is Ellipse with (h,k)=(3,1),a=4,b=sqrt(17)