What is ((x-3)^2)/(18)+((y+4)^2)/(32)=1 ?

The solution to ((x-3)^2)/(18)+((y+4)^2)/(32)=1 is Ellipse with (h,k)=(3,-4),a=3sqrt(2),b=4sqrt(2)

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((x-3)^2)/(18)+((y+4)^2)/(32)=1

\frac{( x - 3 ) ^{2}}{18} + \frac{( y + 4 ) ^{2}}{32} = 1

(h, k) = (3, - 4), a = 32, b = 42

Solution steps

\frac{( x - 3 ) ^{2}}{18} + \frac{( y + 4 ) ^{2}}{32} = 1

Rewrite

\frac{( x - 3 ) ^{2}}{18} + \frac{( y + 4 ) ^{2}}{32} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (3, - 4), a = 32, b = 42

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (3, - 4), b = 42, a = 32

\frac{( y - 4 ) ^{2}}{9} + \frac{( x - 1 ) ^{2}}{4} = 1

\frac{x ^{2}}{85} + \frac{y ^{2}}{4} = 1

x, \frac{x ^{2}}{16} + \frac{Y ( x ) ^{2}}{4} = 1

\frac{x ^{2} - 2 x + y ^{2}}{4 - y} = 14

X^{2} + 2 Y^{2} - 2 X - 4 Y - 8 = 0

What is ((x-3)^2)/(18)+((y+4)^2)/(32)=1 ?
The solution to ((x-3)^2)/(18)+((y+4)^2)/(32)=1 is Ellipse with (h,k)=(3,-4),a=3sqrt(2),b=4sqrt(2)