Chapter 9.6, Problem 3OE

To determine

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

Answer to Problem 3OE

New equation of ellipse is $\frac{x^2}{25}+\frac{{\left(y+5\right)}^2}{4}=1$

Explanation of Solution

Given information: Equation of conic is $\frac{x^2}{25}+\frac{y^2}{4}=1$ and new center is $\left(0,-5\right)$

Formula Used:

Standard form of the equation of ellipse with center $\left(h,k\right)$ and major axis parallel to x-axis 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

When, $a>b$ , foci lies on x-axis

When $b>a$ , foci lies on y-axis

Calculation:

Equation of conic is given as follows:

  $\frac{x^2}{25}+\frac{y^2}{4}=1$

Rewriting the equation as follows:

  $\frac{x^2}{5^2}+\frac{y^2}{2^2}=1$

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

  $\begin{array}{l}a=5\\ b=2\\ \left(h,k\right)=\left(0,0\right)\end{array}$

Since $a>b$ , thus foci lies on x-axis.

Since center is moved to $\left(0,-5\right)$

Thus, $\left(h,k\right)=\left(0,-5\right)$

New equation of ellipse is:

  $\begin{array}{l}\frac{{\left(x-0\right)}^2}{5^2}+\frac{{\left(y-\left(-5\right)\right)}^2}{2^2}=1\\ \frac{x^2}{25}+\frac{{\left(y+5\right)}^2}{4}=1\end{array}$

Conclusion:

Hence, new equation of ellipse is $\frac{x^2}{25}+\frac{{\left(y+5\right)}^2}{4}=1$