What is x^2-16x+4y^2-12y+16=0 ?

The solution to x^2-16x+4y^2-12y+16=0 is Ellipse with (h,k)=(8, 3/2),a=sqrt(57),b=(sqrt(57))/2

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x^2-16x+4y^2-12y+16=0

x^{2} - 16 x + 4 y^{2} - 12 y + 16 = 0

(h, k) = (8, \frac{3}{2}), a = 57, b = \frac{57}{2}

Solution steps

x^{2} - 16 x + 4 y^{2} - 12 y + 16 = 0

Rewrite

x^{2} - 16 x + 4 y^{2} - 12 y + 16 = 0

in the form of the standard ellipse equation

\frac{( x - 8 ) ^{2}}{( 57 ) ^{2}} + \frac{( y - \frac{3}{2} ) ^{2}}{( \frac{57}{2} ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (8, \frac{3}{2}), a = 57, b = \frac{57}{2}

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (8, \frac{3}{2}), a = 57, b = \frac{57}{2}

5 x^{2} + 3 y^{2} - 15 = 0

\frac{y ^{2}}{b ^{2}} = 1 - \frac{x ^{2}}{a ^{2}}

- 1 (- 4 x^{2} - y^{2} + 16) \geq - 1 (0)

\frac{x ^{2}}{14} + \frac{y ^{2}}{8} = 1

- 1.14 y^{2} = 52.3 + x^{2}

What is x^2-16x+4y^2-12y+16=0 ?
The solution to x^2-16x+4y^2-12y+16=0 is Ellipse with (h,k)=(8, 3/2),a=sqrt(57),b=(sqrt(57))/2