Chapter 9.6, Problem 3E

To determine

To find:

The equation of the parabola in the form of ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$

Answer to Problem 3E

  $\frac{{\left(x+1\right)}^2}{10}+\frac{{\left(y-3\right)}^2}{10}=1$

Explanation of Solution

Given:

The two point on the parabola that is $-1,3$

Concept used:

Formula for the parabola

  ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$

If $h=-1,k=3,r=\sqrt{10}$ , this equation reduces to ${\left(x\right)}^2+{\left(y\right)}^2=r^2$

Where is an equation of a parabola of radius $r$ with centre at the origin?

Calculation:

D is on the parabola the radius r is $d\left(C,D\right)$

Using the standard equation of the parabola with

  ${\left(x+1\right)}^2+{\left(y-3\right)}^2={\left(r\right)}^2$

  $r=\sqrt{10}$ ,

An equation of a parabola of radius $r$

Using these value

  $\begin{array}{l}{\left(x+1\right)}^2+{\left(y-3\right)}^2={\left(\sqrt{10}\right)}^2\\ \frac{{\left(x+1\right)}^2}{10}+\frac{{\left(y-3\right)}^2}{10}=1\end{array}$