Chapter 9.6, Problem 5OE

To determine

To calculate: To identify the conic and to find the center and foci

Answer to Problem 5OE

New equation of conic is $\frac{{\left(x+4\right)}^2}{49}-\frac{{\left(y-3\right)}^2}{49}=1$

Explanation of Solution

Given information: Equation of conic is $x^2-y^2=49$ and new center is $\left(-4,3\right)$

Formula Used:

Standard form of the equation of hyperbola with center $\left(h,k\right)$ is

  $\frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}=1$

Where, $a=$ x-intercept, $b=$ y-intercept

Calculation:

Equation of conic is given as follows:

  $x^2-y^2=49$

Rewriting the equations:

  $\begin{array}{l}\frac{x^2}{49}-\frac{y^2}{49}=1\\ \frac{x^2}{7^2}-\frac{y^2}{7^2}=1\end{array}$

Comparing the equation with standard form of the equation of ellipse $\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1$ :

  $\left(h,k\right)=\left(0,0\right)$

Since center is moved to $\left(-4,3\right)$

Thus, $\left(h,k\right)=\left(-4,3\right)$

New equation of ellipse is:

  $\begin{array}{l}\frac{{\left(x-\left(-4\right)\right)}^2}{49}-\frac{{\left(y-3\right)}^2}{49}=1\\ \frac{{\left(x+4\right)}^2}{49}-\frac{{\left(y-3\right)}^2}{49}=1\end{array}$

Conclusion:

Hence, new equation of conic is $\frac{{\left(x+4\right)}^2}{49}-\frac{{\left(y-3\right)}^2}{49}=1$