Let's rewrite the given equation in order to identify the conic section. We will do it by completing the square.

x^{2} + 4 y^{2} - 6 x + 16 y - 11 = 0

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Rewrite equation

Rewrite $- 11$ as $9 + 16 - 36$

x^{2} + 4 y^{2} - 6 x + 16 y + 9 + 16 - 36 = 0

Commutative Property of Addition

x^{2} - 6 x + 9 + 4 y^{2} + 16 y + 16 - 36 = 0

$LHS + 36 = RHS + 36$

x^{2} - 6 x + 9 + 4 y^{2} + 16 y + 16 = 36

Factor out $4$

x^{2} - 6 x + 9 + 4 (y^{2} + 4 y + 4) = 36

$a^{2} \pm 2 a b + b^{2} = (a \pm b)^{2}$

(x - 3)^{2} + 4 (y + 2)^{2} = 36

$LHS / 36 = RHS / 36$

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

Write as a sum of fractions

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

$\frac{a}{b} = \frac{a / 4}{b / 4}$

\frac{( x - 3 ) ^{2}}{3 6} + \frac{( y + 2 ) ^{2}}{9} = 1

Write as a power

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

$a + b = a - (- b)$

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

We can see that the binomials containing the variables are both raised to the power of

2

and are both positive. Moreover, the denominator of the binomial containing the

x -

variable is greater than the denominator of the binomial that contains the

y -

variable. Therefore, our equation matches the format of a horizontal ellipse.

Let's recall the main characteristics of this type of ellipse.

Exercise