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
+128408 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
+128408 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
