Chapter 9.6, Problem 4CYU

To determine

To Find: Standard form and graph of the given equation.

Answer to Problem 4CYU

Parabola; $x=6{\left(y-2\right)}^2+4$

Explanation of Solution

Given:

  $6y^2-24y+28-x=0$

Formula Used:

  $\begin{array}{l}\left(\text{Hyperbola equation : }{B}^2-4AC>0\right)\text{, }\left(\text{Ellipse equation : }{B}^2-4AC<0\text{ and }A\ne C\right)\\ \left(\text{Equation of parabola : }{B}^2-4AC=0\right)\\ \text{STANDARD}\text{}\text{ FORM OF}\text{}\text{}\text{EQUATION }\\ \text{Parabolic equation : }x=a{\left(y-k\right)}^2+h\\ \text{Ellipse equation : }\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\\ \text{Hyperbola equation with centre }\left(h,k\right)\text{ : }\frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}\text{=1 }\\ \text{Circle equation with centre }\left(h,k\right)\text{ : }{\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2\end{array}$

Calculation:

  $\begin{array}{l}6y^2-24y+28-x=0\\ \Rightarrow 6y^2-24y+28=x\\ \Rightarrow y=\frac{-\left(-24\right)}{2\left(6\right)}=\frac{24}{12}=2\left(\text{using }y=\frac{-b}{2a}\text{ and }b=2,a=6\right)\\ \Rightarrow x=6{\left(2\right)}^2-24\left(2\right)+28=4\left(\text{putting y=2 for getting }x\right)\\ \Rightarrow x=6{\left(y-2\right)}^2+4\left(\text{In the form of parabolic equation }x=a{\left(y-k\right)}^2+h\right)\\ \end{array}$

Conclusion:

The given conic is parabola because it has only one variable that is squared.