What is ((x-3)^2)/(169)+((y+4)^2)/(144)=1 ?

The solution to ((x-3)^2)/(169)+((y+4)^2)/(144)=1 is Ellipse with (h,k)=(3,-4),a=13,b=12

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((x-3)^2)/(169)+((y+4)^2)/(144)=1

\frac{( x - 3 ) ^{2}}{169} + \frac{( y + 4 ) ^{2}}{144} = 1

(h, k) = (3, - 4), a = 13, b = 12

Solution steps

\frac{( x - 3 ) ^{2}}{169} + \frac{( y + 4 ) ^{2}}{144} = 1

Rewrite

\frac{( x - 3 ) ^{2}}{169} + \frac{( y + 4 ) ^{2}}{144} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (3, - 4), a = 13, b = 12

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (3, - 4), a = 13, b = 12

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\frac{( x - 2.1 ) ^{2}}{2.25} + \frac{( y + 1.6 ) ^{2}}{6.16} = 1

x^{2} - 2 x + 3 y^{2} - 2 y - 5 = 0

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

2 y - x^{2} - y^{2} \geq 0

What is ((x-3)^2)/(169)+((y+4)^2)/(144)=1 ?
The solution to ((x-3)^2)/(169)+((y+4)^2)/(144)=1 is Ellipse with (h,k)=(3,-4),a=13,b=12