Chapter 4.6, Problem 61S

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 61S

  $y=x^2-1$

Given:

The two point on the parabola that is $C\left(0,0\right)$ and $P\left(0,1\right)$

Concept used:

Formula for the parabola

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

If $h=0,k=0,r=1$ , 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 $d\left(C,D\right)$

By the distance formula

  $r=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

  $\begin{array}{l}r=\sqrt{{\left(0-0\right)}^2+{\left(0-1\right)}^2}\\ r=1\\ r=1\end{array}$

Using the standard equation of the parabola with

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

  $r=1$ ,

An equation of a parabola of radius $r$

Using these value

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

Explanation of Solution

Given:

The two point on the parabola that is $C\left(0,0\right)$ and $P\left(0,1\right)$

Concept used:

Formula for the parabola

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

If $h=0,k=0,r=1$ , 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 $d\left(C,D\right)$

By the distance formula

  $r=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

  $\begin{array}{l}r=\sqrt{{\left(0-0\right)}^2+{\left(0-1\right)}^2}\\ r=1\\ r=1\end{array}$

Using the standard equation of the parabola with

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

  $r=1$ ,

An equation of a parabola of radius $r$

Using these value

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