What is 16x^2+25y^2+224+250y-191=0 ?

The solution to 16x^2+25y^2+224+250y-191=0 is Ellipse with (h,k)=(0,-5),a=sqrt(37),b=(4sqrt(37))/5

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16x^2+25y^2+224+250y-191=0

16 x^{2} + 25 y^{2} + 224 + 250 y - 191 = 0

(h, k) = (0, - 5), a = 37, b = \frac{4 37}{5}

Solution steps

16 x^{2} + 25 y^{2} + 224 + 250 y - 191 = 0

Rewrite

16 x^{2} + 25 y^{2} + 224 + 250 y - 191 = 0

in the form of the standard ellipse equation

\frac{( x - 0 ) ^{2}}{( 37 ) ^{2}} + \frac{( y - ( - 5 ) ) ^{2}}{( \frac{4 37}{5} ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (0, - 5), a = 37, b = \frac{4 37}{5}

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (0, - 5), a = 37, b = \frac{4 37}{5}

\frac{( x + 3 ) ^{2}}{16} + \frac{( y + 2 ) ^{2}}{9} = 1

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

16 x^{(2)} + 25 y^{(2)} - 128 x + 100 y = 0

\frac{( x + 2 ) ^{2}}{64} + \frac{( y - 8 ) ^{2}}{36} = 1

16 x^{2} + 25 y^{2} = \frac{169}{4}

What is 16x^2+25y^2+224+250y-191=0 ?
The solution to 16x^2+25y^2+224+250y-191=0 is Ellipse with (h,k)=(0,-5),a=sqrt(37),b=(4sqrt(37))/5