Chapter 10, Problem 1CT

To determine

The identification of the equation $\frac{{(x+1)}^2}{4}-\frac{y^2}{9}=1$ . If it is a parabola, give a vertex, focus, and directrix; if it is an ellipse, give its center, vertices, and foci; if it is a hyperbola, give its center, vertices, foci, and asymptotes.

Answer to Problem 1CT

Solution:

The equation $\frac{{(x+1)}^2}{4}-\frac{y^2}{9}=1$ represents hyperbola with center $\left(-1,0\right)$, vertices are $(1,0)$ and $(-3,0)$, foci are $\left(-1+\sqrt{13},0\right)$ and $\left(-1-\sqrt{13},0\right)$, the asymptotes are $y=\frac{3}{2}\left(x+1\right)$ and $y=-\frac{3}{2}\left(x+1\right)$ .

Explanation of Solution

Given information:

The equation $\frac{{(x+1)}^2}{4}-\frac{y^2}{9}=1$

The equation $\frac{{(x+1)}^2}{4}-\frac{y^2}{9}=1$ can be rewritten as $\frac{{\left(x-\left(-1\right)\right)}^2}{2^2}-\frac{{\left(y-0\right)}^2}{3^2}=1$ .

Thus, the equation $\frac{{\left(x-\left(-1\right)\right)}^2}{2^2}-\frac{{\left(y-0\right)}^2}{3^2}=1$ is of the form $\frac{{(x-h)}^2}{a^2}-\frac{{(y-k)}^2}{b^2}=1$ .

Hence, $\frac{{(x+1)}^2}{4}-\frac{y^2}{9}=1$ represents the equation of the hyperbola.

The equation of the hyperbola is of the form $\frac{{(x-h)}^2}{a^2}-\frac{{(y-k)}^2}{b^2}=1$ with center at $\left(h,k\right)$, vertices are $\left(h\pm a,k\right)$, asymptotes are $y-k=\pm \frac{b}{a}(x-h)$ and foci $\left(h\pm c,k\right)$, where $c^2=a^2+b^2$ .

Compare the equation with $\frac{{(x-h)}^2}{a^2}-\frac{{(y-k)}^2}{b^2}=1$

  $h=-1,k=0,$

  $a=2$ and $b=3$

So, $c^2=a^2+b^2=4+9=13$

  $\Rightarrow c=\sqrt{13}$

The center of the given equation is $\left(-1,0\right)$ and the transverse axis is the $x$ - axis.

The vertices $\left(h\pm a,k\right)$ are $\left(-1+2,0\right)=(1,0)$ and $\left(-1-2,0\right)=(-3,0)$

The foci $\left(h\pm c,k\right)$ are $\left(-1+\sqrt{13},0\right)$ and $\left(-1-\sqrt{13},0\right)$ .

The asymptotes $y-k=\pm \frac{b}{a}(x-h)$ are $y-0=\pm \frac{3}{2}(x-(-1))$

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

Thus, the equation $\frac{{(x+1)}^2}{4}-\frac{y^2}{9}=1$ represents the hyperbola with center $\left(-1,0\right)$, vertices are $(1,0)$ and $(-3,0)$, foci are $\left(-1+\sqrt{13},0\right)$ and $\left(-1-\sqrt{13},0\right)$, the asymptotes are $y=\frac{3}{2}\left(x+1\right)$ and $y=-\frac{3}{2}\left(x+1\right)$ .