What is 25(x+1)^2+16(y-3)^2=400 ?

The solution to 25(x+1)^2+16(y-3)^2=400 is Ellipse with (h,k)=(-1,3),a=4,b=5

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25(x+1)^2+16(y-3)^2=400

25 (x + 1)^{2} + 16 (y - 3)^{2} = 400

(h, k) = (- 1, 3), a = 4, b = 5

Solution steps

25 (x + 1)^{2} + 16 (y - 3)^{2} = 400

Rewrite

25 (x + 1)^{2} + 16 (y - 3)^{2} = 400

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (- 1, 3), a = 4, b = 5

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (- 1, 3), b = 5, a = 4

9 y^{2} + 90 y + 4 x^{2} - 16 x + 205 = 0

49 x^{2} + 81 y^{2} = 3969

16 x^{2} - 32 x + 7 y^{2} - 33 - 42 y = 0

\frac{( x - 1 ) ^{2}}{5} + \frac{( y - 2 ) ^{2}}{8} = 1

11 x^{2} + 2 y^{2} - 22 = 0

What is 25(x+1)^2+16(y-3)^2=400 ?
The solution to 25(x+1)^2+16(y-3)^2=400 is Ellipse with (h,k)=(-1,3),a=4,b=5