What is ((x+1)^2)/(32)+((y+5)^2)/(36)=1 ?

The solution to ((x+1)^2)/(32)+((y+5)^2)/(36)=1 is Ellipse with (h,k)=(-1,-5),a=4sqrt(2),b=6

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((x+1)^2)/(32)+((y+5)^2)/(36)=1

\frac{( x + 1 ) ^{2}}{32} + \frac{( y + 5 ) ^{2}}{36} = 1

(h, k) = (- 1, - 5), a = 42, b = 6

Solution steps

\frac{( x + 1 ) ^{2}}{32} + \frac{( y + 5 ) ^{2}}{36} = 1

Rewrite

\frac{( x + 1 ) ^{2}}{32} + \frac{( y + 5 ) ^{2}}{36} = 1

in the form of the standard ellipse equation

\frac{( x - ( - 1 ) ) ^{2}}{( 4 2 ) ^{2}} + \frac{( y - ( - 5 ) ) ^{2}}{6 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- 1, - 5), a = 42, b = 6

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (- 1, - 5), b = 6, a = 42

\frac{( x - 6 ) ^{2}}{122} + \frac{( y - 8 ) ^{2}}{\frac{12200}{121}} = 1

\frac{x ^{2}}{3} + \frac{y ^{2}}{2} = 1

x^{2} + 16 y^{2} + 4 x - 32 y - 44 = 0

16 x^{2} + 2 y^{2} - 64 x + 150 y - 11 = 0

25 x^{2} + 16 y^{2} = 1

What is ((x+1)^2)/(32)+((y+5)^2)/(36)=1 ?
The solution to ((x+1)^2)/(32)+((y+5)^2)/(36)=1 is Ellipse with (h,k)=(-1,-5),a=4sqrt(2),b=6