Chapter 3, Problem 8CA

(a)

To determine

To find: The values for the missing data in the given equation.

Answer to Problem 8CA

The required equation is $x=\frac{1}{4}{\left(y+1\right)}^2+1$ .

Explanation of Solution

Given equation:

  $x=\frac{1}{4}{\left(y-k\right)}^2+h$

Consider two quadratic equations:

  $x=\frac{1}{4}{\left(y-k\right)}^2+h$

For a parabola with its vertex in the second quadrant:

k should be negative and h should be positive.

Consider the number to be $1$

Substitute $-1$ for k and $1$ for h in the above equation:

  $x=\frac{1}{4}{\left(y+1\right)}^2+1$

Hence, the required equation is $x=\frac{1}{4}{\left(y+1\right)}^2+1$ .

(b)

To determine

To find: The values for the missing data in the given equation.

Answer to Problem 8CA

The required equation is $x=\frac{1}{4}{\left(y-1\right)}^2+1$ .

Explanation of Solution

Given equation:

  $x=\frac{1}{4}{\left(y-k\right)}^2+h$

Consider two quadratic equations:

  $x=\frac{1}{4}{\left(y-k\right)}^2+h$

For a parabola with its focus at first quadrant:

k should be positive and h should be positive.

Consider the number to be

  $1$

Substitute $1$ for k and $1$ for h in the above equation:

  $x=\frac{1}{4}{\left(y-1\right)}^2+1$

Hence, the required equation is $x=\frac{1}{4}{\left(y-1\right)}^2+1$ .

(c)

To determine

To find: The values for the missing data in the given equation.

Answer to Problem 8CA

The required equation is $x=\frac{1}{4}{\left(y+10\right)}^2-10$ .

Explanation of Solution

Given equation:

  $x=\frac{1}{4}{\left(y-k\right)}^2+h$

Consider two quadratic equations:

  $x=\frac{1}{4}{\left(y-k\right)}^2+h$

For a parabola with its focus at the third quadrant:

k should be negative and h should be negative.

Consider the number to be $10$

Substitute $-1$ for k and $1$ for h in the above equation:

  $x=\frac{1}{4}{\left(y+10\right)}^2-10$

Hence, the required equation is $x=\frac{1}{4}{\left(y+10\right)}^2-10$ .