What is (x+2)^2+32(y-4)^2=16 ?

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

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(x+2)^2+32(y-4)^2=16

(x + 2)^{2} + 32 (y - 4)^{2} = 16

(h, k) = (- 2, 4), a = 4, b = \frac{1}{2}

Solution steps

(x + 2)^{2} + 32 (y - 4)^{2} = 16

Rewrite

(x + 2)^{2} + 32 (y - 4)^{2} = 16

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (- 2, 4), a = 4, b = \frac{1}{2}

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (- 2, 4), a = 4, b = \frac{1}{2}

\frac{( x - 3 ) ^{2}}{16} + \frac{( y - 5 ) ^{2}}{25} = 1

\frac{( x + 5 ) ^{2}}{100} + \frac{( y - 2 ) ^{2}}{36} = 1

\frac{y ^{2}}{36} + \frac{x ^{2}}{100} = 1

\frac{( x - 2 ) ^{2}}{4} + \frac{( y + 3 ) ^{2}}{9} = 1

(x^{2} + 9 x) + (3 y^{2} + 10 y) = 2

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