What is ((x+2)^2)/9+((y-4)^2)/(16)=1 ?

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

Popular >

((x+2)^2)/9+((y-4)^2)/(16)=1

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

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

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

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

25 x^{2} + 64 y^{2} = 1600

\frac{( x + 4 ) ^{2}}{9} + \frac{( y - 6 ) ^{2}}{25} = 1

\frac{( y - 8 ) ^{2}}{40} + \frac{( x + 10 ) ^{2}}{211} = 1

4 y^{2} + 16 x^{2} - 36 x - 20 y + 36 = 0

\frac{x ^{2}}{25} + \frac{y ^{2}}{76} = 1

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