What is ((x-4)^2)/2+((y-3)^2)/1 =1 ?

The solution to ((x-4)^2)/2+((y-3)^2)/1 =1 is Ellipse with (h,k)=(4,3),a=sqrt(2),b=1

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((x-4)^2)/2+((y-3)^2)/1 =1

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

(h, k) = (4, 3), a = 2, b = 1

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (4, 3), a = 2, b = 1

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (4, 3), a = 2, b = 1

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

x^{2} + y^{2} + 8 y = 0

9 - x^{2} - 9 y^{2} > 0

9 x^{2} + 16 y^{2} < 25

4 x^{2} + y^{2} - 8 x + 4 y + 4 = 0

What is ((x-4)^2)/2+((y-3)^2)/1 =1 ?
The solution to ((x-4)^2)/2+((y-3)^2)/1 =1 is Ellipse with (h,k)=(4,3),a=sqrt(2),b=1