What is (2(x-6)^2)/3+(7(y+4)^2)/4 =1 ?

The solution to (2(x-6)^2)/3+(7(y+4)^2)/4 =1 is Ellipse with (h,k)=(6,-4),a=sqrt(3/2),b= 2/(sqrt(7))

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(2(x-6)^2)/3+(7(y+4)^2)/4 =1

\frac{2 ( x - 6 ) ^{2}}{3} + \frac{7 ( y + 4 ) ^{2}}{4} = 1

(h, k) = (6, - 4), a = \frac{3}{2}, b = \frac{2}{7}

Solution steps

\frac{2 ( x - 6 ) ^{2}}{3} + \frac{7 ( y + 4 ) ^{2}}{4} = 1

Rewrite

\frac{2 ( x - 6 ) ^{2}}{3} + \frac{7 ( y + 4 ) ^{2}}{4} = 1

in the form of the standard ellipse equation

\frac{( x - 6 ) ^{2}}{( \frac{3}{2} ) ^{2}} + \frac{( y - ( - 4 ) ) ^{2}}{( \frac{2}{7} ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (6, - 4), a = \frac{3}{2}, b = \frac{2}{7}

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (6, - 4), a = \frac{3}{2}, b = \frac{2}{7}

x^{(2)} + 2 y^{(2)} + 2 x - 20 y + 43 = 0

\frac{( x - 2 ) ^{2}}{16} + \frac{( y + 1 ) ^{2}}{4} = 1

\frac{x ^{2}}{27} + \frac{y ^{2}}{36} = 1

x^{2} + 4 y^{2} \leq 16

16 x^{2} + y^{2} - 32 x + 4 y - 44 = 0

What is (2(x-6)^2)/3+(7(y+4)^2)/4 =1 ?
The solution to (2(x-6)^2)/3+(7(y+4)^2)/4 =1 is Ellipse with (h,k)=(6,-4),a=sqrt(3/2),b= 2/(sqrt(7))