Chapter 9: Q. 21 (page 772)

Complete the square to describe the conics in Exercises 18–21 .
$y^{2} - 8 y - 4 x^{2} - 8 x - 13 = 0$

Short Answer

Expert verified

The answer is $\frac{(y - 4)^{2}}{25} - \frac{4 (x + 1)^{2}}{25} = 1 {H y p e r b o l a}$ with center $(- 1, 4)$ that opens up and down.

Step by step solution

Step 1. Given Information.

The equation is $y^{2} - 8 y - 4 x^{2} - 8 x - 13 = 0$ .

Step 2. Completing the square.

Now,

$(y^{2} - 8 y) - (4 x^{2} + 8 x) - 13 = 0 (y^{2} - 8 y) - 4 (x^{2} + 2 x) - 13 = 0 (y^{2} - 8 y + 16 - 16) - 4 (x^{2} + 2 x + 1 - 1) - 13 = 0 (y^{2} - 8 y + 16) - 16 - 4 (x^{2} + 2 x + 1) + 4 - 13 = 0 (y^{2} - 8 y + 16) - 4 (x^{2} + 2 x + 1) - 25 = 0 (y - 4)^{2} - 4 {(x^{2} + 1)}^{2} - 25 = 0$

Step 3. Final equation.

On simplifying the above equation,

$(y - 4)^{2} - 4 {(x^{2} + 1)}^{2} - 25 + 25 = 0 + 25 (y - 4)^{2} - 4 {(x^{2} + 1)}^{2} = 25 \frac{(y - 4)^{2} - 4 (x + 1)^{2}}{25} = \frac{25}{25} \frac{(y - 4)^{2} - 4 (x + 1)^{2}}{25} = 1 \frac{(y - 4)^{2}}{25} - \frac{4 (x + 1)^{2}}{25} = 1$

which is a hyperbola with center $(- 1, 4)$ that opens up and down.

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Complete the square to describe the conics in Exercises 18–21 .
$y^{2} - 8 y - 4 x^{2} - 8 x - 13 = 0$

Short Answer

Step by step solution

Step 1. Given Information.

Step 2. Completing the square.

Step 3. Final equation.

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Complete the square to describe the conics in Exercises 18–21 .y2-8y-4x2-8x-13=0

Short Answer

Step by step solution

Step 1. Given Information.

Step 2. Completing the square.

Step 3. Final equation.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Pure Maths

Statistics

Applied Mathematics

Calculus

Logic and Functions

Probability and Statistics

Study anywhere. Anytime. Across all devices.

Complete the square to describe the conics in Exercises 18–21 .
$y^{2} - 8 y - 4 x^{2} - 8 x - 13 = 0$