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Intermediate Algebra

OPENSTAX

$Math Studyset 91影视 Explanations$ Math

OER 2017 Edition 路 1346 pages

ISBN: 9780998625720

Chapter 11: Q. 294 (page 1158)

In the following exercises,
(a) write the equation in standard form and
(b) graph.
$25 x^{2} + 4 y^{2} + 150 x + 125 = 0$

Short Answer

Expert verified

The equation in standard form is $\frac{{(x + 3)}^{2}}{4} + \frac{y^{2}}{25} = 1$ and the graph is

Step by step solution

01

Step 1. Given Information

The given equation is $25 x^{2} + 4 y^{2} + 150 x + 125 = 0$

02

Part (a) Step 1. Calculation

From the given equation it is very clear that $x^{2} a n d y^{2}$ term have the same sign and different coefficient. So, the equation represents an ellipse.

Now, group the term and make the coefficient of $x^{2} a n d y^{2}$ equal to 1 and then complete the square and write as binomial using formula.

localid="1645293131461" $(25 x^{2} + 150 x) + 4 y^{2} = - 125 25 (x^{2} + 6 x + 9) + 4 {(y)}^{2} = - 125 + 225 25 {(x + 3)}^{2} + 4 {(y)}^{2} = 100 \frac{25 {(x + 3)}^{2} + 4 {(y)}^{2}}{100} = \frac{100}{100} \frac{{(x + 3)}^{2}}{4} + \frac{y^{2}}{25} = 1$

03

Part (b) Step 1. Graph

On comparing the obtained equation with standard form, we get,

$h = - 3, k = 0$

So, center of ellipse is $(- 3, 0)$

As $25 > 4$ and 25 is in the $y^{2}$ term so major axis is vertical.

The distance from center to vertices is as follows,

$a^{2} = 25 a = 卤 \sqrt{25} a = 卤 5$

And distance from center to endpoint of minor axis is as folllows,

$b^{2} = 4 b = 卤 \sqrt{4} b = 卤 2$

Thus, graph of ellipse is

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Most popular questions from this chapter

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