Chapter 12, Problem 2CR

(a)

To determine

To graph: The circle $x^2+y^2=100$ and parabola $y=3x^2$.

Explanation of Solution

Given information:

The equation of circle is $x^2+y^2=100$ and the equation of parabola is $y=3x^2$.

Graph:

The equation $x^2+y^2=100$ is a circle with radius $10$ and center is origin $\left(0,0\right)$.

Hence, the points on the circle are $\left(-10,0\right),\left(0,-10\right),\left(10,0\right)$ and $\left(0,10\right)$.

Now consider the parabola $y=3x^2$. It represents an upward parabola.

The points on the parabola $y=3x^2$ are shown in the below table.

$\begin{array}{ccc}x& y& \left(x,y\right)\\ -1& 3& \left(-1,3\right)\\ 0& 0& \left(0,0\right)\\ 1& 3& \left(1,3\right)\end{array}$

The graph of a circle $x^2+y^2=100$ and the parabola $y=3x^2$ is,

Interpretation:

The graph represents the parabola $y=3x^2$ and circle $x^2+y^2=100$.

(b)

To determine

The solution of the system of equations $\{\begin{array}{c}{x}^2+y^2=100\\ y=3x^2\end{array}$.

Answer to Problem 2CR

Solution:

The solutions of system of equations $\{\begin{array}{c}{x}^2+y^2=100\\ y=3x^2\end{array}$ are $\left\{\left(\sqrt{\frac{-1+\sqrt{3601}}{18}},\frac{-1+\sqrt{3601}}{6}\right),\left(-\sqrt{\frac{-1+\sqrt{3601}}{18}},\frac{-1+\sqrt{3601}}{6}\right)\right\}$

Explanation of Solution

Given information:

The given system of equations is: $\{\begin{array}{c}{x}^2+y^2=100\\ y=3x^2\end{array}$

Formula used:

For a quadratic equation of the form $a{x}^2+bx+c=0$, the solutions are $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$.

Calculation:

Consider the system of equations,

$x^2+y^2=100$ ….. (1)

$y=3x^2\Rightarrow -3x^2+y=0$ …….. (2)

Multiply equation (1) by $3$,

$\Rightarrow 3x^2+3y^2=300$ ….. (3)

Add equations(2) and (3),

$+\begin{array}{c}3x^2+3y^2=300\\ -3x^2+y=0\\ 3y^2+y=300\end{array}$

$\Rightarrow 3y^2+y-300=0$.

Here, $a=3,b=1$ and $c=-300$.

Substituting these values in the quadratic formula,

$y=\frac{-1\pm \sqrt{1^2-4\left(3\right)\left(-300\right)}}{2\left(3\right)}=\frac{-1\pm \sqrt{3601}}{6}$

Substituting these values of $y$ and solving for $x$,

$y=\frac{-1+\sqrt{3601}}{6}\Rightarrow \frac{-1+\sqrt{3601}}{6}=3x^2\Rightarrow \frac{-1+\sqrt{3601}}{18}=x^2\Rightarrow x=\pm \sqrt{\frac{-1+\sqrt{3601}}{18}}$

$y=\frac{-1-\sqrt{3601}}{6}\Rightarrow \frac{-1-\sqrt{3601}}{6}=3x^2\Rightarrow \frac{-1-\sqrt{3601}}{18}=x^2\Rightarrow x=\pm \sqrt{\frac{-1-\sqrt{3601}}{18}}$

These two solutions are undefined since $\frac{-1-\sqrt{3601}}{18}<0$.

Therefore, the system has solutions $x=\sqrt{\frac{-1+\sqrt{3601}}{18}},y=\frac{-1+\sqrt{3601}}{6}$ and $x=-\sqrt{\frac{-1+\sqrt{3601}}{18}},y=\frac{-1+\sqrt{3601}}{6}$.

Hence, the solutions are $\left\{\left(\sqrt{\frac{-1+\sqrt{3601}}{18}},\frac{-1+\sqrt{3601}}{6}\right),\left(-\sqrt{\frac{-1+\sqrt{3601}}{18}},\frac{-1+\sqrt{3601}}{6}\right)\right\}$

(c)

To determine

The points of intersection for circle $x^2+y^2=100$ and parabola $y=3x^2$.

Answer to Problem 2CR

Solution:

The points of intersection for circle $x^2+y^2=100$ and parabola $y=3x^2$ are $\left(\sqrt{\frac{-1+\sqrt{3601}}{18}},\frac{-1+\sqrt{3601}}{6}\right)\approx \left(1.81,9.84\right)$ and $\left(-\sqrt{\frac{-1+\sqrt{3601}}{18}},\frac{-1+\sqrt{3601}}{6}\right)\approx \left(-1.81,9.84\right)$.

Explanation of Solution

Given information:

The equation of circle is $x^2+y^2=100$ and the equation of parabola is $y=3x^2$.

Formula used:

For a quadratic equation of the form $a{x}^2+bx+c=0$, the solutions are $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$.

Calculation:

From part (b),

The solutions of the system of equations $\{\begin{array}{c}{x}^2+y^2=100\\ y=3x^2\end{array}$ are $\left\{\left(\sqrt{\frac{-1+\sqrt{3601}}{18}},\frac{-1+\sqrt{3601}}{6}\right),\left(-\sqrt{\frac{-1+\sqrt{3601}}{18}},\frac{-1+\sqrt{3601}}{6}\right)\right\}$.

Therefore, the graphs of circle and parabola intersect at these points.

Hence, the point of intersection of circle and parabola are $\left(\sqrt{\frac{-1+\sqrt{3601}}{18}},\frac{-1+\sqrt{3601}}{6}\right)\approx \left(1.81,9.84\right)$ and $\left(-\sqrt{\frac{-1+\sqrt{3601}}{18}},\frac{-1+\sqrt{3601}}{6}\right)\approx \left(-1.81,9.84\right)$.