What is ((x-4)^2}{10}+\frac{(y-2)^2)/6 =1 ?

The solution to ((x-4)^2}{10}+\frac{(y-2)^2)/6 =1 is Ellipse with (h,k)=(4,2),a=sqrt(10),b=sqrt(6)

Popular >

((x-4)^2}{10}+\frac{(y-2)^2)/6 =1

\frac{( x - 4 ) ^{2}}{10} + \frac{( y - 2 ) ^{2}}{6} = 1

(h, k) = (4, 2), a = 10, b = 6

Solution steps

\frac{( x - 4 ) ^{2}}{10} + \frac{( y - 2 ) ^{2}}{6} = 1

Rewrite

\frac{( x - 4 ) ^{2}}{10} + \frac{( y - 2 ) ^{2}}{6} = 1

in the form of the standard ellipse equation

\frac{( x - 4 ) ^{2}}{( 10 ) ^{2}} + \frac{( y - 2 ) ^{2}}{( 6 ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (4, 2), a = 10, b = 6

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (4, 2), a = 10, b = 6

9 x^{2} - 18 x + 4 y^{2} - 27 = 0

4 x^{2} + 25 y^{2} + 8 x = 96

(\frac{x ^{2}}{16}) + (\frac{y ^{2}}{25}) = 1

\frac{( x + 4 ) ^{2}}{6 ^{2}} + \frac{( y + 6 ) ^{2}}{7 ^{2}} = 1

X^{2} + 4 Y^{2} - 6 X - 16 Y + 21 = 0

What is ((x-4)^2}{10}+\frac{(y-2)^2)/6 =1 ?
The solution to ((x-4)^2}{10}+\frac{(y-2)^2)/6 =1 is Ellipse with (h,k)=(4,2),a=sqrt(10),b=sqrt(6)