What is ((x+5)^2)/(100)+((y-4)^2)/(64)=1 ?

The solution to ((x+5)^2)/(100)+((y-4)^2)/(64)=1 is Ellipse with (h,k)=(-5,4),a=10,b=8

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((x+5)^2)/(100)+((y-4)^2)/(64)=1

\frac{( x + 5 ) ^{2}}{100} + \frac{( y - 4 ) ^{2}}{64} = 1

(h, k) = (- 5, 4), a = 10, b = 8

Solution steps

\frac{( x + 5 ) ^{2}}{100} + \frac{( y - 4 ) ^{2}}{64} = 1

Rewrite

\frac{( x + 5 ) ^{2}}{100} + \frac{( y - 4 ) ^{2}}{64} = 1

in the form of the standard ellipse equation

\frac{( x - ( - 5 ) ) ^{2}}{1 0 ^{2}} + \frac{( y - 4 ) ^{2}}{8 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- 5, 4), a = 10, b = 8

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (- 5, 4), a = 10, b = 8

2 x^{2} + 7 y^{2} = 14

\frac{( x - 2 ) ^{2}}{36} + \frac{( y + 1 ) ^{2}}{25} = 1

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

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

\frac{( x + 4 ) ^{2}}{25} + \frac{( y - 2 ) ^{2}}{9} = 1

What is ((x+5)^2)/(100)+((y-4)^2)/(64)=1 ?
The solution to ((x+5)^2)/(100)+((y-4)^2)/(64)=1 is Ellipse with (h,k)=(-5,4),a=10,b=8