What is -1=(((x-0))/(1^2))^2+(((y-1))/(1^2))^2 ?

The solution to -1=(((x-0))/(1^2))^2+(((y-1))/(1^2))^2 is Ellipse with (h,k)=(0,1),a=1,b=1

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-1=(((x-0))/(1^2))^2+(((y-1))/(1^2))^2

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

(h, k) = (0, 1), a = 1, b = 1

Solution steps

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

Rewrite

- 1 - (\frac{x - 0}{1 ^{2}})^{2} - (\frac{y - 1}{1 ^{2}})^{2} = 0

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (0, 1), a = 1, b = 1

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (0, 1), b = 1, a = 1

\frac{u + v}{u ^{2} + v ^{2}} + \frac{- v + u}{u ^{2} + v ^{2}} = 1

12 (x + 4)^{2} + 3 (y - 1)^{2} = 48

\frac{( x + 2 ) ^{2}}{36} + \frac{( y + 5 ) ^{2}}{9} = 1

\frac{x ^{2}}{9} + \frac{y ^{2}}{4} = 1 \cdot \frac{1}{9} \cdot \frac{1}{4}

16 x^{2} + 9 y^{2} - 32 x + 18 y - 119 = 0

What is -1=(((x-0))/(1^2))^2+(((y-1))/(1^2))^2 ?
The solution to -1=(((x-0))/(1^2))^2+(((y-1))/(1^2))^2 is Ellipse with (h,k)=(0,1),a=1,b=1