What is ((x+4)^2)/(64)+((y+2)^2)/(100)=1 ?

The solution to ((x+4)^2)/(64)+((y+2)^2)/(100)=1 is Ellipse with (h,k)=(-4,-2),a=8,b=10

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((x+4)^2)/(64)+((y+2)^2)/(100)=1

\frac{( x + 4 ) ^{2}}{64} + \frac{( y + 2 ) ^{2}}{100} = 1

(h, k) = (- 4, - 2), a = 8, b = 10

Solution steps

\frac{( x + 4 ) ^{2}}{64} + \frac{( y + 2 ) ^{2}}{100} = 1

Rewrite

\frac{( x + 4 ) ^{2}}{64} + \frac{( y + 2 ) ^{2}}{100} = 1

in the form of the standard ellipse equation

\frac{( x - ( - 4 ) ) ^{2}}{8 ^{2}} + \frac{( y - ( - 2 ) ) ^{2}}{1 0 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- 4, - 2), a = 8, b = 10

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (- 4, - 2), b = 10, a = 8

490 x^{(2)} + 10 y^{(2)} + 808 x - 556 y + 40 = 0

16 X^{2} + 25 Y^{2} - 64 X - 50 Y - 311 = 0

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

\frac{x ^{2}}{2.3 2 ^{2}} + \frac{y ^{2}}{2.8 2 ^{2}} = 1

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

What is ((x+4)^2)/(64)+((y+2)^2)/(100)=1 ?
The solution to ((x+4)^2)/(64)+((y+2)^2)/(100)=1 is Ellipse with (h,k)=(-4,-2),a=8,b=10