Q. 21 - Calculus

Question

Complete the square to describe the conics in Exercises 18–21 .

$y^2-8y-4x^2-8x-13=0$

Step-by-Step Solution

Answer

The answer is $\frac{(y-4{)}^2}{25}-\frac{4(x+1{)}^2}{25}=1 \left\{Hyperbola\right\}$ with center $(-1,4)$ that opens up and down.

Step 1. Given Information.

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

Step 2. Completing the square.

Now,

$$\begin{gathered} \left(y^2-8y\right)-\left(4x^2+8x\right)-13=0 \\ \left(y^2-8y\right)-4\left(x^2+2x\right)-13=0 \\ \left(y^2-8y+16-16\right)-4\left(x^2+2x+1-1\right)-13=0 \\ \left(y^2-8y+16\right)-16-4\left(x^2+2x+1\right)+4-13=0 \\ \left(y^2-8y+16\right)-4\left(x^2+2x+1\right)-25=0 \\ (y-4{)}^2-4{\left(x^2+1\right)}^2-25=0 \end{gathered}$$

Step 3. Final equation.

On simplifying the above equation,

$$\begin{gathered} (y-4{)}^2-4{\left(x^2+1\right)}^2-25+25=0+25 \\ (y-4{)}^2-4{\left(x^2+1\right)}^2=25 \\ \frac{(y-4{)}^2-4(x+1{)}^2}{25}=\frac{25}{25} \\ \frac{(y-4{)}^2-4(x+1{)}^2}{25}=1 \\ \frac{(y-4{)}^2}{25}-\frac{4(x+1{)}^2}{25}=1 \end{gathered}$$

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