Chapter 2, Problem 13CR

a

To determine

To write: An equation of the parabola in vertex form.

Answer to Problem 13CR

The equation of the parabola with the vertex $(10,-4)$ and passes through $(1,12)$ in vertex form is $y=\frac{16}{81}{(x-10)}^2-4$ .

Explanation of Solution

Given information: Vertex $(10,-4)$ and passes through $(1,12)$ .

Vertex form of a parabola is $y=a{(x-h)}^2+k$ .

Vertex $(h,k)$ is $(10,-4)$ .

By substituting the vertex in the vertex form of equation we get,

  $y=a{(x-10)}^2-4$

To find the value of $a$ , substitute $(1,12)$ in the above equation.

  $\begin{array}{}$ 12=a{(1-10)}^2-4$$ 12=a{(9)}^2-4$$ 12=a(81)-4$$ 16=a(81)$$ a=\frac{16}{81}$\end{array}$

Substitute $a=\frac{16}{81}$ in $y=a{(x-10)}^2-4$ ,

  $y=\frac{16}{81}{(x-10)}^2-4$

The above equation is the vertex form of the given parabola.

b

To determine

To find: The equation of the parabola with the vertex $(2,6)$ and focus $(2,2)$ .

Answer to Problem 13CR

The equation of the parabola with the vertex $(2,6)$ and focus $(2,2)$ is $y^2=-\frac{1}{16}{(x-2)}^2+6$ .

Explanation of Solution

Given information: Vertex $(2,6)$ and focus $(2,2)$ .

Standard form of the parabola with a vertex and focus is,

  $y^2=\frac{1}{4p}{(x-h)}^2+k$

The vertex is $(h,k)$ and the focus is $(h,k+p)$ , therefore $h=2,k=6,p=-4$ .

  $y^2=\frac{1}{4(-4)}{(x-2)}^2+6$

The above equation is the equation of the parabola with the Vertex $(2,6)$ and focus $(2,2)$ .