What is ((x-3)^2)/(16)+((y+2)^2)/(25)=1 ?

The solution to ((x-3)^2)/(16)+((y+2)^2)/(25)=1 is Ellipse with (h,k)=(3,-2),a=4,b=5

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((x-3)^2)/(16)+((y+2)^2)/(25)=1

\frac{( x - 3 ) ^{2}}{16} + \frac{( y + 2 ) ^{2}}{25} = 1

(h, k) = (3, - 2), a = 4, b = 5

Solution steps

\frac{( x - 3 ) ^{2}}{16} + \frac{( y + 2 ) ^{2}}{25} = 1

Rewrite

\frac{( x - 3 ) ^{2}}{16} + \frac{( y + 2 ) ^{2}}{25} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (3, - 2), a = 4, b = 5

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (3, - 2), b = 5, a = 4

9 x^{2} + 25 y^{2} - 54 x + 100 y - 44 = 0

2 x^{2} - 8 x + y^{2} - 6 y + 7 = 0

3 x^{2} + 2 y^{2} - 12 x + 6 y - 10 = 0

4.14 x^{2} + 3.92 y^{2} - 41.82 x - 82.62 y + 78 = 0

36 x^{2} + 8 y^{2} = 98

What is ((x-3)^2)/(16)+((y+2)^2)/(25)=1 ?
The solution to ((x-3)^2)/(16)+((y+2)^2)/(25)=1 is Ellipse with (h,k)=(3,-2),a=4,b=5