What is ((x-1)^2)/9+((y+2)^2)/(16)=1 ?

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

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((x-1)^2)/9+((y+2)^2)/(16)=1

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

(h, k) = (1, - 2), a = 3, b = 4

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (1, - 2), a = 3, b = 4

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (1, - 2), b = 4, a = 3

144 x^{2} + 169 y^{2} - 288 x + 676 y - 33516 = 0

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

25 x^{2} + 50 x + 4 y^{2} - 16 y = 59

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

3 x^{2} + 8 y^{2} + 12 x + 48 y + 60 = 1

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