Chapter 10, Problem 76RE

To determine

Tofind:the collection of the points in a plane if distance from the point is five- fourths of its distance from the line.

Answer to Problem 76RE

The collection of all points gives a hyperbola and the equation of the hyperbola is $\frac{{\text{x}}^{\text{2}}}{\text{16}}\text{-}\frac{{\text{y}}^{\text{2}}}{\text{9}}\text{=1}$ .

Explanation of Solution

Given:

Distance from the point $\left(\text{5,0}\right)$ .

  ${\text{d}}_{\text{1}}\text{=}\frac{5}{\text{4}}{\text{d}}_{\text{2}}$ .

Distance from the line $\text{x=}\frac{\text{16}}{5}$ .

Concept used:

Distance formula:

  $\text{d=}\sqrt{{\left(\text{x-h}\right)}^{\text{2}}\text{+}{\left(\text{y-k}\right)}^{\text{2}}}$ .

Calculation:

Let any point on the curve be $\left(\text{x,y}\right)$

Distance from the point $\left(\text{5,0}\right)$ .

  ${\text{d}}_1=\sqrt{{\left(x-5\right)}^2+y^2}$ .

Distance from the line $\text{x=}\frac{\text{16}}{5}$ .

  ${\text{d}}_{\text{2}}\text{=x-}\frac{\text{16}}{5}$ .

  ${\text{d}}_{\text{1}}\text{=}\frac{5}{\text{4}}{\text{d}}_{\text{2}}$ .

  $\frac{5}{\text{4}}\left(\text{x-}\frac{\text{16}}{5}\right)\text{=}\sqrt{{\left(\text{x-5}\right)}^{\text{2}}{\text{+y}}^{\text{2}}}$ .

  $\frac{25}{16}{\left(\text{x-}\frac{\text{16}}{5}\right)}^2\text{=}\left({\left(\text{x-5}\right)}^{\text{2}}{\text{+y}}^{\text{2}}\right)$ .

  $\frac{\text{25}}{\text{16}}\left({\text{x}}^{\text{2}}\text{+}\frac{{\text{16}}^{\text{2}}}{\text{25}}\text{-}\frac{\text{32x}}{\text{5}}\right){\text{=x}}^{\text{2}}{\text{+25-10x+y}}^{\text{2}}$ .

  $\left(\text{1-}\frac{\text{25}}{\text{16}}\right){\text{x}}^{\text{2}}{\text{+9+y}}^{\text{2}}\text{=0}$

  $\text{-}\frac{{\text{9x}}^{\text{2}}}{\text{16}}{\text{+y}}^{\text{2}}\text{=-9}$ .

  $\frac{{\text{x}}^{\text{2}}}{\text{16}}\text{-}\frac{{\text{y}}^{\text{2}}}{\text{9}}\text{=1}$ .

Hence, the collection of all points gives a hyperbola and the equation of the hyperbola is $\frac{{\text{x}}^{\text{2}}}{\text{16}}\text{-}\frac{{\text{y}}^{\text{2}}}{\text{9}}\text{=1}$ .