Chapter 10, Problem 38RE

To determine

To find: A description the collection of points in a plane so that the distance from each point to the point $\left(3,0\right)$ is three fourths of its distance from the line $x=16/3$.

Answer to Problem 38RE

Ellipse; $\frac{x^2}{16}+\frac{y^2}{7}=1$

Explanation of Solution

Given:

The distance from each point to the point $\left(3,0\right)$ is three fourths of its distance from the line $x=16/3$.

Calculation:

Let $\left(x,y\right)$ be set of all points.

The distance from $\left(x,y\right)$ to $\left(3,0\right)$ is $\sqrt{{\left(x-3\right)}^2+y^2}$.

The distance from $\left(x,y\right)$ to the line $x=\frac{16}{3}$ is $\left|x-\frac{16}{3}\right|$.

The distance from $\left(3,0\right)$ is three fourths of its distance from the line $x=\frac{16}{3}$.

$\sqrt{{\left(x-3\right)}^2+y^2}=\frac{3}{4}\left|x-\frac{16}{3}\right|$

${\left(x-3\right)}^2+y^2=\frac{9}{16}{\left(x-\frac{16}{3}\right)}^2$

$x^2+9-6x+y^2=\frac{9}{16}\left(x^2+\frac{256}{9}-\frac{32}{3}x\right)$

$7x^2+16y^2=112$

$\frac{7}{112}{x}^2+\frac{16}{112}{y}^2=1$

$\frac{x^2}{16}+\frac{y^2}{7}=1$

Hence, from the above equation we see that the set of points describes an ellipse.