Chapter 11, Problem 36RE

To determine

To find: The type of curve, the focus, vertices and to plot the equation through a graph.

Answer to Problem 36RE

The equation given is a parabola

Co-ordinates of vertices $\left(1,-\frac{13}{6}\right)$

Co-ordinates of focus $\left(1,-\frac{5}{3}\right)$

Explanation of Solution

Given information:

The equation of the curve is:

  $3x^2-6(x+y)=10$

Calculation:

On rearranging the equation, we get a equation of parabola whose vertices are not at origin. Standard equation of parabola when vertices are $(h,k)$ is given by:

  ${(x-h)}^2=4a(y-k)\to 1$

Here axis of symmetry is $x=h$ and focus = $\left(h,a+k\right)$

  $3x^2-6(x+y)=10$

Rearrange the above equation,

  $\begin{array}{l}3x^2-6x=6y+10\\ (x^2-2x)=(2y+\frac{10}{3})\\ (x^2-2x+1-1)=(2y+\frac{10}{3})\\ {(}^x=2y+\frac{10}{3}+1\end{array}$

  $\begin{array}{l}{(}^x=2y+\frac{13}{3}\\ {(}^x=2\left(y+\frac{13}{6}\right)\to 2\end{array}$

Comparing equation $1$ and $2$

Vertices $=\left(1,-\frac{13}{6}\right)$

Focus $=1,\left(\frac{1}{2}-\frac{13}{6}\right)$

  $\begin{array}{l}=\left(1,-\frac{10}{6}\right)\\ =\left(1,-\frac{5}{3}\right)\end{array}$

Graph:

  

Interpretation :

The equation given is a parabola

Co-ordinates of vertices $\left(1,-\frac{13}{6}\right)$

Co-ordinates of focus $\left(1,-\frac{5}{3}\right)$