What is ((x-4)^2)/(16)+(y^2)/(40.96)=1 ?

The solution to ((x-4)^2)/(16)+(y^2)/(40.96)=1 is Ellipse with (h,k)=(4,0),a=4,b=sqrt(40.96)

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((x-4)^2)/(16)+(y^2)/(40.96)=1

\frac{( x - 4 ) ^{2}}{16} + \frac{y ^{2}}{40.96} = 1

(h, k) = (4, 0), a = 4, b = 40.96

Solution steps

\frac{( x - 4 ) ^{2}}{16} + \frac{y ^{2}}{40.96} = 1

Rewrite

\frac{( x - 4 ) ^{2}}{16} + \frac{y ^{2}}{40.96} = 1

in the form of the standard ellipse equation

\frac{( x - 4 ) ^{2}}{4 ^{2}} + \frac{( y - 0 ) ^{2}}{( 40.96 ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (4, 0), a = 4, b = 40.96

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (4, 0), b = 40.96, a = 4

2 x^{2} + 2 y^{2} - 28 y = 0

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

6 x^{2} + 25 y^{2} - 160 x - 300 y - 300 = 0

x^{2} + \frac{y ^{2}}{49} = 1

\frac{( x + 2 ) ^{2}}{49} + \frac{( y - 3 ) ^{2}}{4} = 1

What is ((x-4)^2)/(16)+(y^2)/(40.96)=1 ?
The solution to ((x-4)^2)/(16)+(y^2)/(40.96)=1 is Ellipse with (h,k)=(4,0),a=4,b=sqrt(40.96)