What is ((x-5)^2)/(12)+((y+2)^2)/(16)=1 ?

The solution to ((x-5)^2)/(12)+((y+2)^2)/(16)=1 is Ellipse with (h,k)=(5,-2),a=2sqrt(3),b=4

Popular >

((x-5)^2)/(12)+((y+2)^2)/(16)=1

\frac{( x - 5 ) ^{2}}{12} + \frac{( y + 2 ) ^{2}}{16} = 1

(h, k) = (5, - 2), a = 23, b = 4

Solution steps

\frac{( x - 5 ) ^{2}}{12} + \frac{( y + 2 ) ^{2}}{16} = 1

Rewrite

\frac{( x - 5 ) ^{2}}{12} + \frac{( y + 2 ) ^{2}}{16} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (5, - 2), a = 23, b = 4

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (5, - 2), b = 4, a = 23

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

x^{2} + 2 y^{2} - 10 x + 8 y + 29 = 0

16 x^{2} + 4 y^{2} - 32 x + 16 y - 32 = 0

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

\frac{( x - 5 ) ^{2}}{49} + \frac{( y + 4 ) ^{2}}{81} = 1

What is ((x-5)^2)/(12)+((y+2)^2)/(16)=1 ?
The solution to ((x-5)^2)/(12)+((y+2)^2)/(16)=1 is Ellipse with (h,k)=(5,-2),a=2sqrt(3),b=4