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Chapter 10: Q48. (page 677)

In Problems 43–52, identify the graph of each equation without applying a rotation of axes.
$10 x^{2} + 12 x y + 4 y^{2} - x - y + 10 = 0$

Short Answer

Expert verified

The equation defines ellipse.

Step by step solution

01

Step 1. Given Information

The given equation is $10 x^{2} + 12 x y + 4 y^{2} - x - y + 10 = 0$

Here $A = 10, B = 12, C = 4$

02

Step 2. Identifying Conic

$B^{2} - 4 A C = 12^{2} - 4 (10) (4) B^{2} - 4 A C = 144 - 160 = - 16 < 0$

The equation defines ellipse.

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Most popular questions from this chapter

The graph of a parabola is given. Match graph to its equation.

and equations of the parabola are $A) y^{2} = 4 x C) y^{2} = - 4 x E) {(y - 1)}^{2} = 4 (x - 1) G) {(y - 1)}^{2} = - 4 (x - 1) B) x^{2} = 4 y D) x^{2} = - 4 y F) {(x + 1)}^{2} = 4 (y + 1) H) {(x + 1)}^{2} = 4 (y + 1)$

To graph $y = {(x - 3)}^{2} + 1$ , shift the graph of $y = x^{2}$ to the right________ units and then________ 1 unit.

In Problems 43–52, identify the graph of each equation without applying a rotation of axes. $x^{2} + 3 x y - 2 y^{2} + 3 x + 2 y + 5 = 0$

True or False

The equation $a x^{2} + 6 y^{2} - 12 y = 0$ defines an ellipse if width="36">a>0

Find the equation of the parabola described. Find the two points that define the latus rectum, and graph the equation by hand.

Vertex is at $(4, - 2)$ and focus is at $(6, - 2)$ .

See all solutions

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