Chapter 10.2, Problem 10AYU

To determine

To find:

The equation of the parabola is of the given form

Answer to Problem 10AYU

  $y^2-16x-2y+52=0$

Explanation of Solution

Given:

The vertex

  $V=\left(3,2\right)$

Concept used:

The stabdard form of parabola equation

  $y=a{x}^2+bx+c$

Putting $x=h,y=k$ in equation (1)

  $k=a{h}^2+bh+c\mathrm{.......................}\left(2\right)$

Where $\left(h,k\right)$ is the vertex

Calculation:

The stabdard form of parabola equation

  $y=a{x}^2+bx+c$

Putting $x=h,y=k$ in equation (1)

  $\begin{array}{l}k=a{h}^2+bh+c\\ {\left(y-k\right)}^2=4a\left(x-h\right)\mathrm{.......................}\left(2\right)\end{array}$

Where $\left(h,k\right)$ is the vertex

Given the vertex $V=\left(3,2\right)$

Putting in equation (2)

  $\begin{array}{l}{\left(y-2\right)}^2=4a\left(x-3\right)\\ y^2+4-2y=4ax-12a\\ y^2-4ax-2y+12a+4=0\end{array}$

Putting the value of $a=4$

  $\begin{array}{l}{y}^2-4ax-2y+12a+4=0\\ y^2-16x-2y+52=0\end{array}$