What is ((x-3)^2)/(25)+((y-4)^2)/(36)=1 ?

The solution to ((x-3)^2)/(25)+((y-4)^2)/(36)=1 is Ellipse with (h,k)=(3,4),a=5,b=6

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((x-3)^2)/(25)+((y-4)^2)/(36)=1

\frac{( x - 3 ) ^{2}}{25} + \frac{( y - 4 ) ^{2}}{36} = 1

(h, k) = (3, 4), a = 5, b = 6

Solution steps

\frac{( x - 3 ) ^{2}}{25} + \frac{( y - 4 ) ^{2}}{36} = 1

Rewrite

\frac{( x - 3 ) ^{2}}{25} + \frac{( y - 4 ) ^{2}}{36} = 1

in the form of the standard ellipse equation

\frac{( x - 3 ) ^{2}}{5 ^{2}} + \frac{( y - 4 ) ^{2}}{6 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (3, 4), a = 5, b = 6

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (3, 4), b = 6, a = 5

2 x^{2} + y^{2} + 16 x - 4 y + 32 = 0

\frac{( x - 5 ) ^{2}}{16} + \frac{( y - 3 ) ^{2}}{4} = 1

4 y^{2} - 338 x + 32 y = - 169 x^{2} + 443

2 x^{2} + y^{2} - 4 x + 8 y - 4 = 0

25 x^{2} + 9 y^{2} - 250 x - 18 y + 409 = 0

What is ((x-3)^2)/(25)+((y-4)^2)/(36)=1 ?
The solution to ((x-3)^2)/(25)+((y-4)^2)/(36)=1 is Ellipse with (h,k)=(3,4),a=5,b=6