Chapter 9, Problem 15CR

To determine

To find: The equation of the ellipse.

Answer to Problem 15CR

The equation of the ellipse is $3x^2+4y^2=12x$

Explanation of Solution

Given:

Sum of focal radii is $4$

Calculation:

Since

  $\begin{array}{l}2a=4\\ a=2\end{array}$

The sum of focal radii of an ellipse is represented as $2a$

From the given foci, the center of the ellipse is not at the origin. Let the center be $(h,k)$ . Since the second coordinate of each focus is the same, it is an ellipse with horizontal axis. The standard form of the equation of such an ellipse is,

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

The foci is at $(h-c,k)and(h+c,k)$ , where $c^2=a^2-b^2$

The foci of the ellipse $(h-c,k)and(h+c,k)$ are given as $(1,0)and(3,0)$ . These points give us the value of $has2,kas0,andcas1.$

Thus the center of the ellipse is halfway between the foci at $(2,0)$ , and the distance from the center to each focus is $1$ .

Find the distance $b$ using the formula $b^2=a^2-c^2.$

Replace $a$ with $2$ , and $c$ with $1$ in formula.

  $b^2=2^2-1^2$

Simplify.

  $\begin{array}{l}{b}^2=4-1\\ =3\end{array}$

Take square root on both the sides.

  $\begin{array}{l}\sqrt{b}=\sqrt{3}\\ b=3\end{array}$

Substitute $2forh,0fork,\sqrt{3}forb,and2$ for $a$ in the standard form.

  $\begin{array}{l}\frac{{\left(x-2\right)}^2}{2^2}+\frac{{\left(y-0\right)}^2}{{(\sqrt{3})}^2}=1\\ \frac{{\left(x-2\right)}^2}{4}+\frac{y^2}{3}=1\end{array}$

Multiply both sides of the equation by the least common denominator, $12$

  $\begin{array}{l}\left[\frac{{\left(x-2\right)}^2}{4}\right](12)+\left(\frac{y^2}{3}\right)(12)=1(12)\\ 3{\left(x-2\right)}^2+4y^2=12\end{array}$

Simplify.

  $\begin{array}{l}3\left(x^2-4x+4\right)+4y^2=12\\ 3x^2-12x+12+4y^2=12\end{array}$

Add $12x$ and subtract $12$ from both the sides.

  $\begin{array}{l}3x^2-12x+12+4y^2+12x-12=12+12x-12\\ 3x^2+4y^2=12x\end{array}$

Conclusion:

Therefore, the equation of the ellipse is $3x^2+4y^2=12x$

The correct answer is choice $d.$