What is ((x-3)^2}{16}+\frac{(y+4)^2)/9 =1 ?

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

Popular >

((x-3)^2}{16}+\frac{(y+4)^2)/9 =1

\frac{( x - 3 ) ^{2}}{16} + \frac{( y + 4 ) ^{2}}{9} = 1

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

Solution steps

\frac{( x - 3 ) ^{2}}{16} + \frac{( y + 4 ) ^{2}}{9} = 1

Rewrite

\frac{( x - 3 ) ^{2}}{16} + \frac{( y + 4 ) ^{2}}{9} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

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

9 x^{2} + 4 y^{2} + 18 x - 16 y - 11 = 0

3 x^{2} + 2 y^{2} - 12 x + 8 y + 19 = 0

\frac{( x + 1 ) ^{2}}{4} + \frac{( y - 5 ) ^{2}}{9} = 1

\frac{( x + 5 ) ^{2}}{16} + \frac{( y - 3 ) ^{2}}{25} = 1

\frac{y ^{2}}{16} + \frac{x ^{2}}{9} = 1

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