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

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

Popular >

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

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

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

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

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

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

9 (x + 2)^{2} + 4 (y + 1)^{2} = 1

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

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