Question

Describe 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$x=\frac{16}{3}$.

Short answer

The collection of points is the ellipse$\frac{x^2}{16}+\frac{y^2}{7}=1$

Step-by-step solution

Step 1. Given Information

Given to describe 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 $x=\frac{16}{3}$.

Step 2. Distance between the points

Let the point required in the collection be $\left(x,y\right)$.

So the distance between the points $\left(x,y\right)$and $\left(3,0\right)$is:

$$\begin{gathered} d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2} \\ d=\sqrt{{\left(x-3\right)}^2+{\left(y-0\right)}^2} \\ d=\sqrt{{\left(x-3\right)}^2+y^2} \end{gathered}$$

Step 3. Distance to the line

The distance between the point $\left(x,y\right)$and the line $x=\frac{16}{3}$is:

$d=\left|x-\frac{16}{3}\right|$

Step 4. Calculation

Given the distance to the point is three-fourths of the distance to the line.

Hence,

$$\begin{gathered} \sqrt{{\left(x-3\right)}^2+y^2}=\frac{3}{4}\left|x-\frac{16}{3}\right| \\ {\left(x-3\right)}^2+y^2=\frac{9}{16}{\left(x-\frac{16}{3}\right)}^2 \\ {x}^2-6x+9+y^2=\frac{9}{16}{x}^2-6x+16 \\ \frac{7}{16}{x}^2+y^2=7 \\ \frac{x^2}{16}+\frac{y^2}{7}=1 \end{gathered}$$

Step 5. Collection of points

So the collection of points is represented by the ellipse $\frac{x^2}{16}+\frac{y^2}{7}=1$where the major axis is along the x-axis and the minor axis is along the y-axis and the center lies at the origin. The vertices along the major axis are $\left(-4,0\right),\left(4,0\right)$and the vertices along the minor axis are$\left(0,-\sqrt{7}\right),\left(0,\sqrt{7}\right)$. The foci of the ellipse are$\left(-3,0\right),\left(3,0\right)$.

Step 6. Conclusion

The collection of points is the ellipse$\frac{x^2}{16}+\frac{y^2}{7}=1$.