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Conside two fixed point A(0,-2) and B (0,4) and the  line  L is y=1

Consider a moving point Q such that QA is always perpendicular to QB.

1.The equation of Q is ? ,Denote the locus Q

2.Another moving point R is always equidistant form the point S(0.6) nd the line L 

Find the equation of locus R and denote locus R

Guest Mar 9, 2015

Best Answer 

 #1
avatar+81031 
+10

For the first one, let the points A =(0, -2) and B = (0, 4) be endpoints of a diameter of a circle with a radius of 3 centered at (0,1).

Then,  the locus of points formed when QA is perpendicular to QB will lie on this circle and it will have the equation x^2 + (y - 1)^2 = 9. Note that, except at the endoints of the diameter, QA and QB will always from two legs of a right triangle with the diameter AB as a hypotenuse.  {To be technically correct, the points A and B shouldn't actually be included in this locus, since the diameter can't be perpendicular to itself.}

Here's a pic

 

 

For the second, this is just a parabola with a vertex of (0, 7/2) and a width of "a"

The equation is y = ax^2 + (7/2).....here's the graph when a = 1......

https://www.desmos.com/calculator/i3ums3kx9f

 

  

CPhill  Mar 9, 2015
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1+0 Answers

 #1
avatar+81031 
+10
Best Answer

For the first one, let the points A =(0, -2) and B = (0, 4) be endpoints of a diameter of a circle with a radius of 3 centered at (0,1).

Then,  the locus of points formed when QA is perpendicular to QB will lie on this circle and it will have the equation x^2 + (y - 1)^2 = 9. Note that, except at the endoints of the diameter, QA and QB will always from two legs of a right triangle with the diameter AB as a hypotenuse.  {To be technically correct, the points A and B shouldn't actually be included in this locus, since the diameter can't be perpendicular to itself.}

Here's a pic

 

 

For the second, this is just a parabola with a vertex of (0, 7/2) and a width of "a"

The equation is y = ax^2 + (7/2).....here's the graph when a = 1......

https://www.desmos.com/calculator/i3ums3kx9f

 

  

CPhill  Mar 9, 2015

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