What is (x-3)^2+((y+1)^2)/(49)=1 ?

The solution to (x-3)^2+((y+1)^2)/(49)=1 is Ellipse with (h,k)=(3,-1),a=1,b=7

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(x-3)^2+((y+1)^2)/(49)=1

(x - 3)^{2} + \frac{( y + 1 ) ^{2}}{49} = 1

(h, k) = (3, - 1), a = 1, b = 7

Solution steps

(x - 3)^{2} + \frac{( y + 1 ) ^{2}}{49} = 1

Rewrite

(x - 3)^{2} + \frac{( y + 1 ) ^{2}}{49} = 1

in the form of the standard ellipse equation

\frac{( x - 3 ) ^{2}}{1 ^{2}} + \frac{( y - ( - 1 ) ) ^{2}}{7 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (3, - 1), a = 1, b = 7

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (3, - 1), b = 7, a = 1

22 y^{2} + 32 x^{2} - 308 y - 512 x + 2.422 = 0

- 3 y^{2} - 3 x^{2} + 5 x + 10 y = 5 x^{2} + 1

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

(\frac{x - 5}{5})^{2} + (\frac{y + 7}{16})^{2} = 1

2 x^{2} + 4 y^{2} + 18 x + 24 y + 44 = 0

What is (x-3)^2+((y+1)^2)/(49)=1 ?
The solution to (x-3)^2+((y+1)^2)/(49)=1 is Ellipse with (h,k)=(3,-1),a=1,b=7