Chapter 11, Problem 30RE

To determine

To find: The equation of the conic on the given graph.

Answer to Problem 30RE

The equation of the conic is: $\frac{{(x-1)}^2}{I}-\frac{{(y-0)}^2}{I}=1$

Explanation of Solution

Given information: A graph is given which represents a hyperbola $\frac{{(x-h)}^2}{a^2}-\frac{{(y-k)}^2}{b^2}=I$

  

Calculation:

Vertex= $(h\pm a,k)$

Asymptotes= $y-k=\pm \frac{b}{a}(x-h)$

From the given graph;

Centre (h,k) = (1,0) i.e. h=1; k=0

And vertices are; (0,0) and (2,0) (1

Now, we know that;

Vertex= $(h\pm a,k)$ = $(1\pm a,0)$ (2

On comparing 1 and 2,

0=1-a and 2=1+a a=1 and a=1

Similarly, the asymptotes of the ellipse from the graph are:

Asymptote passing through (1,0) and (0,-1) is

  $\begin{array}{l}y-y_1^{}=(\frac{y_2^{}-y_1^{}}{x_2^{}-x_1^{}})(x-x_1^{})\\ =y-0=\frac{-1}{-1}(x-1)\\ y=1(x-1)\end{array}$

Another asymptote passing through (1,0) and (0,1) is:

  $\begin{array}{l}y-y_1^{}=(\frac{y_2^{}-y_1^{}}{x_2^{}-x_1^{}})(x-x_1^{})\\ =y-0=\frac{1}{-1}(x-1)\\ y=-1(x-1)\end{array}$

Now, we know that;

Asymptote; $\begin{array}{l}y-k=\pm \frac{b}{a}(x-h)\\ y-0=\pm \frac{b}{1}(x-1)\\ y=\pm \frac{b}{1}(x-1)\end{array}$

On comparing 3, 4, 5, we get b=1,

Finally, the equation of given conic is

  $\begin{array}{l}\frac{{(x-h)}^2}{a^2}-\frac{{(y-k)}^2}{b^2}=I\\ \frac{{(x-I)}^2}{1^2}-\frac{{(y-k0)}^2}{1^2}=I\\ \frac{{(x-1)}^2}{1}-\frac{{(y-0)}^2}{1}=I\end{array}$

Hence, The equation of the conic is: $\frac{{(x-1)}^2}{I}-\frac{{(y-0)}^2}{I}=1$