What is (x+5)^2+4(y-4)^2=16 ?

The solution to (x+5)^2+4(y-4)^2=16 is Ellipse with (h,k)=(-5,4),a=4,b=2

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(x+5)^2+4(y-4)^2=16

(x + 5)^{2} + 4 (y - 4)^{2} = 16

(h, k) = (- 5, 4), a = 4, b = 2

Solution steps

(x + 5)^{2} + 4 (y - 4)^{2} = 16

Rewrite

(x + 5)^{2} + 4 (y - 4)^{2} = 16

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (- 5, 4), a = 4, b = 2

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (- 5, 4), a = 4, b = 2

(x^{2} - 6 x) + 4 (y^{2} + 2 y) = - 9

\frac{( x + 5 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{25} = 1

\frac{( ( x + 0 ) ^{2} )}{100} + \frac{( ( y + 0 ) ^{2} )}{49} = 1

2 x^{2} + 3 y^{2} + 9 x + 10 y - 3 = 0

\frac{( x - 9 ) ^{2}}{64} + \frac{( y + 2 ) ^{2}}{25} = 1

What is (x+5)^2+4(y-4)^2=16 ?
The solution to (x+5)^2+4(y-4)^2=16 is Ellipse with (h,k)=(-5,4),a=4,b=2