What is (x+6)^2+((y+4)^2)/(1/16)=1 ?

The solution to (x+6)^2+((y+4)^2)/(1/16)=1 is Ellipse with (h,k)=(-6,-4),a=1,b= 1/4

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(x+6)^2+((y+4)^2)/(1/16)=1

(x + 6)^{2} + \frac{( y + 4 ) ^{2}}{\frac{1}{16}} = 1

(h, k) = (- 6, - 4), a = 1, b = \frac{1}{4}

Solution steps

(x + 6)^{2} + \frac{( y + 4 ) ^{2}}{\frac{1}{16}} = 1

Rewrite

(x + 6)^{2} + \frac{( y + 4 ) ^{2}}{\frac{1}{16}} = 1

in the form of the standard ellipse equation

\frac{( x - ( - 6 ) ) ^{2}}{1 ^{2}} + \frac{( y - ( - 4 ) ) ^{2}}{( \frac{1}{4} ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- 6, - 4), a = 1, b = \frac{1}{4}

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (- 6, - 4), a = 1, b = \frac{1}{4}

4 (x - 1)^{2} + 9 y^{2} = 36

\frac{( x ^{2} + 6 x + 9 )}{49} + \frac{( y ^{2} - 2 y + 1 )}{45} = 1

\frac{( y - 5 ) ^{2}}{16} + \frac{( x - 3 ) ^{2}}{4} = 1

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

25 x^{2} + 4 y^{2} + 50 x - 16 y - 59 = 0

What is (x+6)^2+((y+4)^2)/(1/16)=1 ?
The solution to (x+6)^2+((y+4)^2)/(1/16)=1 is Ellipse with (h,k)=(-6,-4),a=1,b= 1/4