What is ((x-5)^2)/9+((y-3)^2)/(16)=1 ?

The solution to ((x-5)^2)/9+((y-3)^2)/(16)=1 is Ellipse with (h,k)=(5,3),a=3,b=4

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((x-5)^2)/9+((y-3)^2)/(16)=1

\frac{( x - 5 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{16} = 1

(h, k) = (5, 3), a = 3, b = 4

Solution steps

\frac{( x - 5 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{16} = 1

Rewrite

\frac{( x - 5 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{16} = 1

in the form of the standard ellipse equation

\frac{( x - 5 ) ^{2}}{3 ^{2}} + \frac{( y - 3 ) ^{2}}{4 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (5, 3), a = 3, b = 4

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (5, 3), b = 4, a = 3

9 x^{2} + 25 y^{2} - 900 = 0

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

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

25 x^{2} + 16 y^{2} + 100 x + 32 y - 284 = 0

4 X^{2} + 8 Y^{2} - 24 X + 32 Y - 12 = 0

What is ((x-5)^2)/9+((y-3)^2)/(16)=1 ?
The solution to ((x-5)^2)/9+((y-3)^2)/(16)=1 is Ellipse with (h,k)=(5,3),a=3,b=4