What is 4X^2+8Y^2-24X+32Y-12=0 ?

The solution to 4X^2+8Y^2-24X+32Y-12=0 is Ellipse with (h,k)=(3,-2),a=2sqrt(5),b=sqrt(10)

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4X^2+8Y^2-24X+32Y-12=0

4 X^{2} + 8 Y^{2} - 24 X + 32 Y - 12 = 0

(h, k) = (3, - 2), a = 25, b = 10

Solution steps

4 X^{2} + 8 Y^{2} - 24 X + 32 Y - 12 = 0

Rewrite

4 X^{2} + 8 Y^{2} - 24 X + 32 Y - 12 = 0

in the form of the standard ellipse equation

\frac{( X - 3 ) ^{2}}{( 2 5 ) ^{2}} + \frac{( Y - ( - 2 ) ) ^{2}}{( 10 ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (3, - 2), a = 25, b = 10

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (3, - 2), a = 25, b = 10

\frac{( x + 2 ) ^{2}}{4} + \frac{( y - 4 ) ^{2}}{20} = 1

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

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

(\frac{x + 1}{2})^{2} + (\frac{y - 1}{4})^{2} = \frac{5}{16}

\frac{( x + 9 ) ^{2}}{2} + \frac{y ^{2}}{3} = 1

What is 4X^2+8Y^2-24X+32Y-12=0 ?
The solution to 4X^2+8Y^2-24X+32Y-12=0 is Ellipse with (h,k)=(3,-2),a=2sqrt(5),b=sqrt(10)