What is ((x-4)^2)/(16)+((y+3)^2)/(25)=1 ?

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

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((x-4)^2)/(16)+((y+3)^2)/(25)=1

\frac{( x - 4 ) ^{2}}{16} + \frac{( y + 3 ) ^{2}}{25} = 1

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

Solution steps

\frac{( x - 4 ) ^{2}}{16} + \frac{( y + 3 ) ^{2}}{25} = 1

Rewrite

\frac{( x - 4 ) ^{2}}{16} + \frac{( y + 3 ) ^{2}}{25} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

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

\frac{( x + 6.5 ) ^{2}}{56.25} + \frac{( y - 3 ) ^{2}}{36} = 1

x^{2} - 4 x + 4 + y^{2} = \frac{x ^{2}}{4} - \frac{16 x}{4} + \frac{64}{4}

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

x^{2} + 12 y^{2} - 12 x - 48 y + 72 = 0

220 x^{2} + 256 y^{2} - 660 x - 3025 = 0

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