Chapter MPS, Problem 10.8MPS

(a)

To determine

To Write: The standard form of the given equation

Answer to Problem 10.8MPS

  $-\frac{1}{80}{\left(x-40\right)}^2+20$

Explanation of Solution

Given:

  $y+\left(\frac{1}{80}\right)x^2-x=0$

Formula Used:

  $\begin{array}{l}A{x}^2+Bxy+c{y}^2+Dx+Ey+F=0\\ if\\ B^2-4AC<0Ellipse\\ ifA=CCircle\\ B^2-4AC=0Parabola\\ B^2-4AC>0Hyperbola\end{array}$

Calculation:

  $\begin{array}{l}y+\left(\frac{1}{80}\right)x^2-x=0\\ \text{⇒}80y+x^2-80x=0\\ \text{⇒}80y+x^2-2\times 40\times x-1600+1600=0\\ \text{⇒}80y+{\left(x-40\right)}^2-1600=0\\ \text{⇒}y=-\frac{1}{80}{\left(x-40\right)}^2+20\end{array}$

Conclusion:

Thus the standard form of given equation is $-\frac{1}{80}{\left(x-40\right)}^2+20$ .

(b)

To determine

To Find: The equation is graph of parabola, hyperbola, ellipse or circle.

Answer to Problem 10.8MPS

Parabola

Explanation of Solution

Given:

  $y+\left(\frac{1}{80}\right)x^2-x=0$

Formula Used:

  $\begin{array}{l}A{x}^2+Bxy+c{y}^2+Dx+Ey+F=0\\ if\\ B^2-4AC<0Ellipse\\ ifA=CCircle\\ B^2-4AC=0Parabola\\ B^2-4AC>0Hyperbola\end{array}$

Calculation:

  $\begin{array}{l}A=1\\ B=0\\ C=0\\ B^2-4ac=0-4\left(1\right)\left(0\right)=0\end{array}$

Conclusion:

Thus the given graph is of parabola.

(c)

To determine

  $y+\left(\frac{1}{80}\right)x^2-x=0$

Explanation of Solution

Given:

  $y+\left(\frac{1}{80}\right)x^2-x=0$

Formula Used:

  $\begin{array}{l}A{x}^2+Bxy+c{y}^2+Dx+Ey+F=0\\ if\\ B^2-4AC<0Ellipse\\ ifA=CCircle\\ B^2-4AC=0Parabola\\ B^2-4AC>0Hyperbola\end{array}$

Calculation:

  $\begin{array}{l}y+\left(\frac{1}{80}\right)x^2-x=0\\ \text{⇒}80y+x^2-80x=0\\ \text{⇒}80y+x^2-2\times 40\times x-1600+1600=0\\ \text{⇒}80y+{\left(x-40\right)}^2-1600=0\\ \text{⇒}y=-\frac{1}{80}{\left(x-40\right)}^2+20\end{array}$

  

Conclusion:

Thus the graph has drawn.