What is ((x-3)^2)/(25)+((y+2)^2)/(49)=1 ?

The solution to ((x-3)^2)/(25)+((y+2)^2)/(49)=1 is Ellipse with (h,k)=(3,-2),a=5,b=7

Popular >

((x-3)^2)/(25)+((y+2)^2)/(49)=1

\frac{( x - 3 ) ^{2}}{25} + \frac{( y + 2 ) ^{2}}{49} = 1

(h, k) = (3, - 2), a = 5, b = 7

Solution steps

\frac{( x - 3 ) ^{2}}{25} + \frac{( y + 2 ) ^{2}}{49} = 1

Rewrite

\frac{( x - 3 ) ^{2}}{25} + \frac{( y + 2 ) ^{2}}{49} = 1

in the form of the standard ellipse equation

\frac{( x - 3 ) ^{2}}{5 ^{2}} + \frac{( y - ( - 2 ) ) ^{2}}{7 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (3, - 2), a = 5, b = 7

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (3, - 2), b = 7, a = 5

16 x^{2} + y^{2} + 96 x - 10 y = - 153

x^{2} + y^{2} - 25 = \frac{x + 25}{7} - y

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

(y - 6)^{2} = 23 - \frac{23 ( x - 3 ) ^{2}}{39}

\frac{x ^{2}}{1.58 7 ^{2}} + \frac{y ^{2}}{0.9 9 ^{2}} = 1

What is ((x-3)^2)/(25)+((y+2)^2)/(49)=1 ?
The solution to ((x-3)^2)/(25)+((y+2)^2)/(49)=1 is Ellipse with (h,k)=(3,-2),a=5,b=7