Chapter 11: Q. 308 (page 1158)

In the following exercises,
(a) write the equation in standard form and
(b) graph.
$36 y^{2} - 16 x^{2} - 96 x + 216 y - 396 = 0$

Short Answer

Expert verified

The equation in standard form is $\frac{{(y + 3)}^{2}}{16} - \frac{{(x + 3)}^{2}}{36} = 1$ and the graph of hyperbola is

Step by step solution

Step 1. Given Information

The given equation is $36 y^{2} - 16 x^{2} - 96 x + 216 y - 396 = 0$

Part(a) Step 1. Calculation

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

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

role="math" localid="1645416233684" $(36 y^{2} + 216 y) - (16 x^{2} + 96 x) = 396 36 (y^{2} + 6 y) - 16 (x^{2} + 6 x) = 396 36 (y^{2} + 6 y + 9) - 16 (x^{2} + 6 x + 9) = 396 - 144 + 324 36 {(y + 3)}^{2} - 16 {(x + 3)}^{2} = 576 \frac{36 {(y + 3)}^{2} - 16 {(x + 3)}^{2}}{576} = \frac{576}{576} \frac{{(y + 3)}^{2}}{16} - \frac{{(x + 3)}^{2}}{36} = 1$

Part(b) Step 1. Calculation and Graph

As $y^{2}$ term is positive, so the transverse axis is vertical.

On comparing with standard form, we get,

$(h, k) = (- 3, - 3)$

And

$a^{2} = 16 a = \sqrt{16} a = 4 A n d b^{2} = 36 b = \sqrt{36} b = 6$

Thus, center is $(- 3, - 3) a n d a = 4, b = 6$

Next, we will find the equation of asymptote as follows,

$y = \frac{a}{b} (x - h) + k y = \frac{4}{6} (x + 3) - 3 y = \frac{2}{3} x + 2 - 3 y = \frac{2}{3} x - 1 A n d y = - \frac{a}{b} (x - h) + k y = - \frac{4}{6} (x + 3) - 3 y = - \frac{2}{3} x - 2 - 3 y = - \frac{2}{3} x - 5$

Thus, the graph of hyperbola is

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In the following exercises,
(a) write the equation in standard form and
(b) graph.
$36 y^{2} - 16 x^{2} - 96 x + 216 y - 396 = 0$

Short Answer

Step by step solution

Step 1. Given Information

Part(a) Step 1. Calculation

Part(b) Step 1. Calculation and Graph

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In the following exercises,(a) write the equation in standard form and(b) graph.36y2-16x2-96x+216y-396=0

Short Answer

Step by step solution

Step 1. Given Information

Part(a) Step 1. Calculation

Part(b) Step 1. Calculation and Graph

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Calculus

Applied Mathematics

Mechanics Maths

Geometry

Discrete Mathematics

Logic and Functions

Study anywhere. Anytime. Across all devices.

In the following exercises,
(a) write the equation in standard form and
(b) graph.
$36 y^{2} - 16 x^{2} - 96 x + 216 y - 396 = 0$