What is 16(x-1)^2+25(y-4)^2=400 ?

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

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16(x-1)^2+25(y-4)^2=400

16 (x - 1)^{2} + 25 (y - 4)^{2} = 400

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

Solution steps

16 (x - 1)^{2} + 25 (y - 4)^{2} = 400

Rewrite

16 (x - 1)^{2} + 25 (y - 4)^{2} = 400

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

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

49 x^{2} + 25 y^{2} = 1225

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

(\frac{4}{9 - x ^{2} - y ^{2}})^{2} = 16

\frac{( y - 8 ) ^{2}}{9} + \frac{( x - 5 ) ^{2}}{4} = 1

0 = 4 x^{2} + y^{2} + 1

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