Chapter 10, Problem 3CT

To determine

The identification of the equation $2x^2+3y^2+4x-6y=13$ . 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 3CT

Solution:

The equation $2x^2+3y^2+4x-6y=13$ represents equation of ellipse, with centerat $\left(-1,1\right)$, vertices are $(2,1)$ and $(-4,1)$, foci are $\left(-1+\sqrt{3},1\right)$ and $\left(-1-\sqrt{3},1\right)$ .

Explanation of Solution

Given information:

The equation $2x^2+3y^2+4x-6y=13$

Consider $2x^2+3y^2+4x-6y=13$

Grouping the terms to complete the squares in $x$ and in $y$ ,

  $\Rightarrow (2x^2+4x)+(3y^2-6y)=13$

  $\Rightarrow 2(x^2+2x)+3(y^2-2y)=13$

  $\Rightarrow 2(x^2+2x+1)+3(y^2-2y+1)=13+2+3$

  $\Rightarrow 2{(x+1)}^2+3{(y-1)}^2=18$

Divide by $18$ on both sides,

  $\Rightarrow \frac{{(x+1)}^2}{9}+\frac{{(y-1)}^2}{6}=1$

Thus, the equation $\frac{{(x+1)}^2}{9}+\frac{{(y-1)}^2}{6}=1$ is of the form $\frac{{(x-h)}^2}{a^2}+\frac{{(y-k)}^2}{b^2}=1$, which is an ellipse.

Hence, $2x^2+3y^2+4x-6y=13$ represents the equation of the ellipse.

The major axis is parallel to $x$ -axis.

The equation of the ellipse is of the form $\frac{{(x-h)}^2}{a^2}+\frac{{(y-k)}^2}{b^2}=1$ ; $a>b>0$ with center at $\left(h,k\right)$, vertices are $\left(h\pm a,k\right)$ and foci are $\left(h\pm c,k\right)$, where $c^2=a^2-b^2$

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

  $h=-1,k=1,$

  $a=3$ and $b=\sqrt{6}$

So, $c^2=a^2-b^2=9-6=3$

  $\Rightarrow c=\sqrt{3}$

The center $\left(h,k\right)$ of the given equation is $\left(-1,1\right)$ .

The vertices $\left(h\pm a,k\right)$ are $\left(-1+3,1\right)=(2,1)$ and $\left(-1-3,1\right)=(-4,1)$

The foci $\left(h\pm c,k\right)$ are $\left(-1+\sqrt{3},1\right)$ and $\left(-1-\sqrt{3},1\right)$ .

Thus, the equation $2x^2+3y^2+4x-6y=13$ represents an ellipse with center $\left(-1,1\right)$, vertices are $(2,1)$ and $(-4,1)$, foci are $\left(-1+\sqrt{3},1\right)$ and $\left(-1-\sqrt{3},1\right)$ .