What is ((x-1)^2)/(425)+((y-2)^2)/(400)=1 ?

The solution to ((x-1)^2)/(425)+((y-2)^2)/(400)=1 is Ellipse with (h,k)=(1,2),a=5sqrt(17),b=20

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((x-1)^2)/(425)+((y-2)^2)/(400)=1

\frac{( x - 1 ) ^{2}}{425} + \frac{( y - 2 ) ^{2}}{400} = 1

(h, k) = (1, 2), a = 517, b = 20

Solution steps

\frac{( x - 1 ) ^{2}}{425} + \frac{( y - 2 ) ^{2}}{400} = 1

Rewrite

\frac{( x - 1 ) ^{2}}{425} + \frac{( y - 2 ) ^{2}}{400} = 1

in the form of the standard ellipse equation

\frac{( x - 1 ) ^{2}}{( 5 17 ) ^{2}} + \frac{( y - 2 ) ^{2}}{2 0 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (1, 2), a = 517, b = 20

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (1, 2), a = 517, b = 20

x^{2} = - y^{2} + 2 y

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

\frac{( x - 1 ) ^{2}}{16} + \frac{( y - 2 ) ^{2}}{49} = 1

4 x^{2} + 16 y^{2} - 16 x - 96 y + 96 = 0

17 x^{2} + 16 y^{2} - 102 x - 32 y - 103 = 0

What is ((x-1)^2)/(425)+((y-2)^2)/(400)=1 ?
The solution to ((x-1)^2)/(425)+((y-2)^2)/(400)=1 is Ellipse with (h,k)=(1,2),a=5sqrt(17),b=20