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

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

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

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

(h, k) = (1, 3), a = \frac{1}{2}, b = \frac{1}{5}

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (1, 3), a = \frac{1}{2}, b = \frac{1}{5}

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (1, 3), a = \frac{1}{2}, b = \frac{1}{5}

(\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

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

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

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