Chapter 11, Problem 1PRE

To determine

To identify each equation whether it represents circle, an ellipse, a hyperbola or parabola.

If it represent the circle identify radius and center.

If the equation represents the ellipse identify the center, vertices, end points, minor axis, foci and eccentricity.

If the equation represents the hyperbola identify the center, vertices, foci, equations of asymptotes and eccentricity.

If the equation represents the ellipse identify the vertex, focus, equation of the directrix, and the equation of the axis of symmetry.

Answer to Problem 1PRE

Center $=(2,-2)$ ,Vertices $=(6,-2);(-2,-2)$ ,Foci $=(7,-2);(-3,-2)$ , the equation of asymptotes: $y=\frac{3}{4}x-\frac{7}{2}$ , $y=-\frac{3}{4}x-\frac{1}{2}$ eccentricity $e=\frac{5}{4}$

Explanation of Solution

Given information:

The given equation is $\frac{{(x-2)}^2}{16}-\frac{{\left(y+2\right)}^2}{9}=1$

Calculation:

Let us consider the given equation-

  $\frac{{(x-2)}^2}{16}-\frac{{\left(y+2\right)}^2}{9}=1$ .

The given equation can be rewritten as follows-

  $\frac{{(x-2)}^2}{4^2}-\frac{{\left(y+2\right)}^2}{3^2}=1$

The given equation is the equation of hyperbola with transverse axis horizontal.

Comparing the given equation with the standard form of equation of hyperbola with transverse axis horizontal is

  $\frac{{(x-h)}^2}{b^2}-\frac{{\left(y-k\right)}^2}{a^2}=1$

Here the center is $(h,k)=(2,-2)$

Vertices:

  $\begin{array}{l}(h+a,k);(h-a,k)\\ =(2+4,-2);(2-4,-2)\\ =(6,-2);(-2,-2)\end{array}$

Foci:

  $\begin{array}{l}(h+c,k);(h-c,k)\\ =(2+5,-2);(2-5,-2)\{c=\sqrt{a^2+b^2}=\sqrt{4^2+3^2}=5\}\\ =(7,-2);(-3,-2)\end{array}$

The formula to find the equation of asymptotes is $y-k=\pm \frac{b}{a}(x-h)$

Substitute the values of variables and solving the given equation as follows-

  $\begin{array}{l}y-k=\frac{b}{a}(x-h)\\ y-(-2)=\frac{3}{4}(x-2)\\ y+2=\frac{3}{4}x-\frac{3}{2}\\ y=\frac{3}{4}x-\frac{3}{2}-2\\ y=\frac{3}{4}x-\frac{7}{2}\end{array}$   $\begin{array}{l}y-k=-\frac{b}{a}(x-h)\\ y-(-2)=-\frac{3}{4}(x-2)\\ y+2=-\frac{3}{4}x+\frac{3}{2}\\ y=-\frac{3}{4}x+\frac{3}{2}-2\\ y=-\frac{3}{4}x-\frac{1}{2}\end{array}$

The eccentricity of the hyperbola is $e=\frac{c}{a}$

  $\begin{array}{l}e=\frac{c}{a}\\ e=\frac{5}{4}\end{array}$