What is (((x+2)^2))/9+(((y-4)^2))/(16)=1 ?

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

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(((x+2)^2))/9+(((y-4)^2))/(16)=1

\frac{( ( x + 2 ) ^{2} )}{9} + \frac{( ( y - 4 ) ^{2} )}{16} = 1

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

Solution steps

\frac{( x + 2 ) ^{2}}{9} + \frac{( y - 4 ) ^{2}}{16} = 1

Rewrite

\frac{( x + 2 ) ^{2}}{9} + \frac{( y - 4 ) ^{2}}{16} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

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

\frac{( x + 1 ) ^{2}}{25} + \frac{( y - 2 ) ^{2}}{9} = 1

- 16 y + 52 = - 2 x^{2} - 8 x - y^{2}

10 x^{2} + 4 y^{2} = 14

\frac{( x + 6 ) ^{2}}{25} + \frac{( y + 1 ) ^{2}}{16} = 1

2 x^{2} - 5 x + 3 y^{2} - 4 y = 1

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