What is ((x-4)^2)/(32)+((y-3)^2)/(36)=1 ?

The solution to ((x-4)^2)/(32)+((y-3)^2)/(36)=1 is Ellipse with (h,k)=(4,3),a=4sqrt(2),b=6

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((x-4)^2)/(32)+((y-3)^2)/(36)=1

\frac{( x - 4 ) ^{2}}{32} + \frac{( y - 3 ) ^{2}}{36} = 1

(h, k) = (4, 3), a = 42, b = 6

Solution steps

\frac{( x - 4 ) ^{2}}{32} + \frac{( y - 3 ) ^{2}}{36} = 1

Rewrite

\frac{( x - 4 ) ^{2}}{32} + \frac{( y - 3 ) ^{2}}{36} = 1

in the form of the standard ellipse equation

\frac{( x - 4 ) ^{2}}{( 4 2 ) ^{2}} + \frac{( y - 3 ) ^{2}}{6 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (4, 3), a = 42, b = 6

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (4, 3), b = 6, a = 42

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\frac{( x + 5 ) ^{2}}{16} + \frac{( y - 4 ) ^{2}}{4} = 1

3 x^{2} + 3 y^{2} - 18 y = 0

1 = 4 x^{2} + 9 y^{2}

X^{2} + 3 Y^{2} - 4 X - 23 = 0

What is ((x-4)^2)/(32)+((y-3)^2)/(36)=1 ?
The solution to ((x-4)^2)/(32)+((y-3)^2)/(36)=1 is Ellipse with (h,k)=(4,3),a=4sqrt(2),b=6