Solve Lesson 84 Page 99 Math 10 – Kite >

Topic

In the coordinate plane Oxygenfor two points A(1 ; 0) and REMOVE(0 ; 3). Find the set of points USA satisfy

GHOST = 2MB.

Solution method – See details

Step 1: Parameterize the coordinates of point M and then calculate the length GHOST, MB

Step 2: Transform the hypothesis GHOST = 2MB then conclude about the set of points USA satisfy

Detailed explanation

Call USA(x ; y)

We have: $$\overrightarrow {AM} = (a – 1;b) \Rightarrow AM = \sqrt {{{(x – 1)}^2} + {y^2}} \Rightarrow A{M^2 } = {(x – 1)^2} + {y^2}$$

$$\overrightarrow {BM} = (a;b – 3) \Rightarrow BM = \sqrt {{x^2} + {{(y – 3)}^2}} \Rightarrow B{M^2} = { x^2} + {(y – 3)^2}$$

By assumption, $$MA = 2MB \Rightarrow M{A^2} = 4M{B^2}$$ $$\Leftrightarrow {(x – 1)^2} + {y^2} = 4\left[ {{x^2} + {{(y – 3)}^2}} \right]$$

$$\Leftrightarrow 3{x^2} + 3{y^2} + 2x – 24y + 35 = 0$$$$\Leftrightarrow {x^2} + {y^2} + \frac{2}{3 }x – 8y + \frac{{35}}{3} = 0$$

$$\Leftrightarrow {\left( {x + \frac{1}{3}} \right)^2} + {\left( {y – 4} \right)^2} = \frac{{40}} {9}$$

So the set of points USA satisfy GHOST = 2MB is a circle with PT: $${\left( {x + \frac{1}{3}} \right)^2} + {\left( {y – 4} \right)^2} = \frac{ {40}}{9}$$ with center $$I\left( { – \frac{1}{3};4} \right)$$ and radius $$R = \frac{{2\sqrt {10} }}{3}$$.