Chapter 9, Problem 26RE

a.

To determine

Calculate the equation of the parabola and the circle.

Answer to Problem 26RE

The equation of parabola and circle are $x^2=-4(y-4)$ and $x^2+{(y+4\sqrt{3})}^2=64$ .

Explanation of Solution

Given:

It is given in the question that a church window is bounded on top by a parabola nad beloe by the arc of a circle.

  

Concept Used:

In this, use the concept of parabola and circle and the equation used will be ${(x-h)}^2=4p(y-k)$ and also use the pythgoras theorem, the equation of circle is ${(x-h)}^2+{(y-k)}^2=r^2$ .

Calculation: Since, from the diagram the parabola has a vertical axis. So, ${(x-h)}^2=4p(y-k)$ .

Now, use the vertex $(0,4)$ and the point $(4,0)$ to find p. Write the equation of the parabola.

  $\begin{array}{l}$ {(x-h)}^2=4p(y-k)$$ {(4-0)}^2=4p(0-4)$16=-16p\\ p=-1\end{array}$

Put the above value in the equation ,

  $\begin{array}{l}$ {(x-h)}^2=4p(y-k)$$ {(x-0)}^2=4(-1)(y-4)$x^2=-4(y-4)\end{array}$

Now, since the x − coordinate of the center of the circle is zero. Use the radius of the circle, the point , and the pythagorean theorem to find y − coordinate . The y- coordinate is negative.

  $\begin{array}{l}{x}^2+y^2=h^2\\ 4^2+y^2=8^2\\ y^2=64-16\\ y^2=\sqrt{48}\\ y=-4\sqrt{3}\end{array}$

So, the equation of circle is

  ${(x-h)}^2+{(y-k)}^2=r^2$

  ${(x-0)}^2+{(y-(-4\sqrt{3}))}^2=8^2$

  $x^2+{(y+4\sqrt{3})}^2=64$

Conclusion:

The equation of parabola and circle are $x^2=-4(y-4)$ and $x^2+{(y+4\sqrt{3})}^2=64$ .

b.

To determine

Find the different values for the table for d.

Answer to Problem 26RE

The all values of d are $2.928,2.741,2.182,1.292,0$ .

Explanation of Solution

Given: It is given in the question that the $d=(\frac{-x^2}{4}+4)-(\sqrt{64-x^2}-4\sqrt{3})$ .

Concept Used:

In this, use the concept of putting the different value of x in d and get the result.

Calculation: Here,

  $\begin{array}{l}d=y(valueofparabola)-y(valueofcircle)\\ d=(\frac{-x^2}{4}+4)-(\sqrt{64-x^2}-4\sqrt{3})\end{array}$

Now,

Put, $x=0$

  $d=(\frac{-0^2}{4}+4)-(\sqrt{64-0^2}-4\sqrt{3})=4-(8-4\sqrt{3})=4-1.072=2.928$

Put, $x=1$

  $d=(\frac{-1^2}{4}+4)-(\sqrt{64-1^2}-4\sqrt{3})=\frac{15}{4}-(\sqrt{63}-4\sqrt{3})=2.741$

Put, $x=2$

  $d=(\frac{-2^2}{4}+4)-(\sqrt{64-2^2}-4\sqrt{3})=\frac{12}{4}-(\sqrt{60}-4\sqrt{3})=3-(2\sqrt{15}-4\sqrt{3})=1.262$

Put, $x=3$

  $d=(\frac{-3^2}{4}+4)-(\sqrt{64-3^2}-4\sqrt{3})=\frac{7}{4}-(\sqrt{55}-4\sqrt{3})=1.262$

Put, $x=4$

  $d=(\frac{-4^2}{4}+4)-(\sqrt{64-4^2}-4\sqrt{3})=0-(\sqrt{48}-4\sqrt{3})=-(4\sqrt{3}-4\sqrt{3})=0$

Conclusion:

The values are $2.928,2.741,2.182,1.292,0$ .