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

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

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

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

(h, k) = (4, 3), a = 1, b = \frac{1}{2}

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (4, 3), a = 1, b = \frac{1}{2}

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (4, 3), a = 1, b = \frac{1}{2}

\frac{x ^{2}}{169} + \frac{y ^{2}}{36} = 1

3 y^{2} + x^{2} - 12 y + 9 = 0

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

16 x^{2} + 4 y^{2} + 16 x + 49 y - 81 = 0

\frac{( x - 4 ) ^{2}}{10} + \frac{( y - 2 ) ^{2}}{6} = 1

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