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

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

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

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

(h, k) = (1, 2), a = 5, b = 4

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (1, 2), a = 5, b = 4

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (1, 2), a = 5, b = 4

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

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

\frac{x}{x ^{2} + y ^{2}} = \frac{3}{2}

4 x^{2} + 25 y^{2} - 50 y - 75 = 0

x^{2} + 4 y^{2} - 2.71 = 0

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