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College Algebra : Ellipses

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Example Question #41 : Graphs

The graph of the equation

$16x^{2}+12y^{2}-96x+48y+384=0$

is an example of which conic section?

Possible Answers:

A horizontal hyperbola

A vertical hyperbola

A horizontal ellipse

A vertical ellipse

The equation has no graph.

Correct answer:

The equation has no graph.

Explanation:

The quadratic coefficients in this general form of a conic equation are 16 and 12. They are of the same sign, making its graph, if it exists, an ellipse.

To determine whether this ellipse is horizontal or vertical, rewrite this equation in standard form

$\frac{\left ( x- h\right )^{2}}{a^{2}} + \frac{\left ( y- k\right )^{2}}{b^{2}} = 1$

as follows:

Subtract 384 from both sides:

$16x^{2}+12y^{2}-96x+48y+384=0$

$16x^{2}+12y^{2}-96x+48y= -384$

Separate the and terms and group them:

$16x^{2} + 12y^{2} -96x-48y =-384$

$(16x^{2} -96x) + (12y^{2}-48y) =-384$

Distribute out the quadratic coefficients:

$16 ( x^{2} -6x ) +12 ( y^{2}-4y) =-384$

Complete the square within each quadratic expression by dividing each linear coefficient by 2 and squaring the quotient.

Since $\left (\frac{-6}{2} \right )^{2}= 9$ and $\left (\frac{-4}{2} \right )^{2}= 4$ , we get

$16 ( x^{2} -6x \textbf{+ 9} ) +12 ( y^{2}-4y\textbf{ + 4}) =-384+ \textbf{?}+ \textbf{?}$

Balance this equation, adjusting for the distributed coefficients:

$16 ( x^{2} -6x + 9 ) +12 ( y^{2}-4y + 4 ) =-384+ 16(9)+ 12(4)$

$16 ( x^{2} -6x + 9 ) +12 ( y^{2}-4y + 4 ) =-192$

The perfect square trinomials are squares of binomials, by design; rewrite them as such:

$16 ( x-3 )^{2} +12 ( y -2 ) ^{2} =-192$

Divide by :

$\frac{16 ( x-3 )^{2}}{-192} +\frac{12 ( y -2 ) ^{2} }{-192}=\frac{-192}{-192}$

$\frac{( x-3 )^{2}}{-12} +\frac{( y -2 ) ^{2} }{-16}=1$

Recall that the standard form of an ellipse is

$\frac{\left ( x- h\right )^{2}}{a^{2}} + \frac{\left ( y- k\right )^{2}}{b^{2}} = 1$

This requires both denominators to be positive. In the standard form of the given equation, they are not. Therefore, the equation has no real ordered pairs as solutions, and it does not have a graph on the coordinate plane.

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Example Question #42 : Graphs

Give the foci of the ellipse of the equation

$\frac{(x- 2)^{2}}{64} + \frac{(y-6)^{2}}{49} = 1$ .

Round your coordinates to the nearest tenth, if applicable.

Possible Answers:

(-1.9, 6), (5.9,6 )

None of the other choices gives the correct response.

(2, -4.6),(2, 16.6)

(-8.6, 6), (12.6, 6)

(2, 2.1), (2, 9.9)

Correct answer:

(-1.9, 6), (5.9,6 )

Explanation:

$\frac{(x- h)^{2}}{a^{2}} + \frac{(y-k)^{2}}{b^{2}} = 1$

is the standard form of an ellipse with center (h,k) . Also, since in the given equation, $a^{2} = 64$ and $b^{2} = 49$ - that is, $a^{2}> b^{2}$ , the ellipse is horizontal.

The foci of a horizontal ellipse are located at

$(h \pm c ,k)$ ,

where

$c= \sqrt{a^{2}-b^{2}} = \sqrt{64-49} = \sqrt{15} \approx 3.9$

Setting $h = 2, k = 6, c \approx 3.9$ , the foci are at

$(2 \pm 3.9, 6)$ , or

(-1.9, 6) and (5.9,6 ) .

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College Algebra : Ellipses

Study concepts, example questions & explanations for College Algebra

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Example Question #41 : Graphs

Example Question #42 : Graphs

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