What is ((x-2)^2}{16}+\frac{(y-3)^2)/9 =1 ?

The solution to ((x-2)^2}{16}+\frac{(y-3)^2)/9 =1 is Ellipse with (h,k)=(2,3),a=4,b=3

Popular >

((x-2)^2}{16}+\frac{(y-3)^2)/9 =1

\frac{( x - 2 ) ^{2}}{16} + \frac{( y - 3 ) ^{2}}{9} = 1

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

Solution steps

\frac{( x - 2 ) ^{2}}{16} + \frac{( y - 3 ) ^{2}}{9} = 1

Rewrite

\frac{( x - 2 ) ^{2}}{16} + \frac{( y - 3 ) ^{2}}{9} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

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

\frac{( x + 3 ) ^{2}}{49} + \frac{( y - 5 ) ^{2}}{16} = 1

18 x^{2} - 72 x + 4 y^{2} + 4 y + 37 = 0

6 (x + 1)^{2} + 2 (y - 1)^{2} = 72

\frac{( x - \frac{5}{2} ) ^{2}}{9} + \frac{y ^{2}}{49} = 1

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

What is ((x-2)^2}{16}+\frac{(y-3)^2)/9 =1 ?
The solution to ((x-2)^2}{16}+\frac{(y-3)^2)/9 =1 is Ellipse with (h,k)=(2,3),a=4,b=3