What is ((X-3)^2)/(169)+((Y-2)^2)/(144)=1 ?

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

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((X-3)^2)/(169)+((Y-2)^2)/(144)=1

\frac{( X - 3 ) ^{2}}{169} + \frac{( Y - 2 ) ^{2}}{144} = 1

(h, k) = (3, 2), a = 13, b = 12

Solution steps

\frac{( X - 3 ) ^{2}}{169} + \frac{( Y - 2 ) ^{2}}{144} = 1

Rewrite

\frac{( X - 3 ) ^{2}}{169} + \frac{( Y - 2 ) ^{2}}{144} = 1

in the form of the standard ellipse equation

\frac{( X - 3 ) ^{2}}{1 3 ^{2}} + \frac{( Y - 2 ) ^{2}}{1 2 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (3, 2), a = 13, b = 12

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (3, 2), a = 13, b = 12

25 x^{2} + 350 x + 16 y^{2} + 96 y - 2231 = 0

\frac{( x - 3 ) ^{2}}{9} + \frac{( y + 4 ) ^{2}}{25} = 1

64 y^{2} - 512 y - (- x^{2} - 16 x + 105) = 0

16 x^{2} + 25 y^{2} - 96 x - 300 y + 644 = 0

4 x^{2} + 36 y^{2} - 144 = 0

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