What is ((x-3)^2)/9+((y+4)^2)/(25)=1 ?

The solution to ((x-3)^2)/9+((y+4)^2)/(25)=1 is Ellipse with (h,k)=(3,-4),a=3,b=5

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((x-3)^2)/9+((y+4)^2)/(25)=1

\frac{( x - 3 ) ^{2}}{9} + \frac{( y + 4 ) ^{2}}{25} = 1

(h, k) = (3, - 4), a = 3, b = 5

Solution steps

\frac{( x - 3 ) ^{2}}{9} + \frac{( y + 4 ) ^{2}}{25} = 1

Rewrite

\frac{( x - 3 ) ^{2}}{9} + \frac{( y + 4 ) ^{2}}{25} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (3, - 4), a = 3, b = 5

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (3, - 4), b = 5, a = 3

64 y^{2} - 512 y - (- x^{2} - 16 x + 105) = 0

16 x^{2} + 25 y^{2} - 96 x - 300 y + 644 = 0

4 x^{2} + 36 y^{2} - 144 = 0

- \frac{6 - x}{4 - y} = - \frac{y}{- 2 - x}

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

What is ((x-3)^2)/9+((y+4)^2)/(25)=1 ?
The solution to ((x-3)^2)/9+((y+4)^2)/(25)=1 is Ellipse with (h,k)=(3,-4),a=3,b=5