What is ((x+3)^2)/(100)+((y+4)^2)/(32)=1 ?

The solution to ((x+3)^2)/(100)+((y+4)^2)/(32)=1 is Ellipse with (h,k)=(-3,-4),a=10,b=4sqrt(2)

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((x+3)^2)/(100)+((y+4)^2)/(32)=1

\frac{( x + 3 ) ^{2}}{100} + \frac{( y + 4 ) ^{2}}{32} = 1

(h, k) = (- 3, - 4), a = 10, b = 42

Solution steps

\frac{( x + 3 ) ^{2}}{100} + \frac{( y + 4 ) ^{2}}{32} = 1

Rewrite

\frac{( x + 3 ) ^{2}}{100} + \frac{( y + 4 ) ^{2}}{32} = 1

in the form of the standard ellipse equation

\frac{( x - ( - 3 ) ) ^{2}}{1 0 ^{2}} + \frac{( y - ( - 4 ) ) ^{2}}{( 4 2 ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- 3, - 4), a = 10, b = 42

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (- 3, - 4), a = 10, b = 42

\frac{x ^{2}}{7} + \frac{y ^{2}}{4} = 1

\frac{( x - 2 ) ^{2}}{5} + \frac{( y + 3 ) ^{2}}{9} = 1

4 x^{2} + 9 y^{2} - 24 x + 18 y + 9 = 0

16 x^{2} + 9 y^{2} - 64 x + 54 y + 1 = 0

3 x^{2} + 3 y^{2} - 30 y = 0

What is ((x+3)^2)/(100)+((y+4)^2)/(32)=1 ?
The solution to ((x+3)^2)/(100)+((y+4)^2)/(32)=1 is Ellipse with (h,k)=(-3,-4),a=10,b=4sqrt(2)