Find the equation of the graph of all points whose distances from ( 0, 3) are 1/3 of their distances from the line x = 27.
anybody please
the distance from (0,3) to the vertical line x = 27 is 27 ... 1/3 of that distance is 9.
now, what do you call the set of all points that are a distance 9 from the point (0,3) , and further ... what is its equation?
also, contrary to what you may think, this is not advanced algebra ... in future, post a problem like this in the Precalculus forum.
My guess is that the word "graph" should be replaced by "locus." More importantly, the question seems to be about points whose distance from (0, 3) is 1/3 of their distance to x = 27, not 1/3 of the distance between (0, 3) and x = 27.
After writing the equation, this seems to be an ellipse according to the classification of conic sections.
Hello, rcs!
Did you make a sketch?
Find the equation of the graph of all points whose distances from (0, 3) are 1/3 of their distances from the line x = 27.We have point $\displaystyle P(x,y)$Code:| | : | (x,y) :B | o * * * * * o(27,y) | * P : A| * : (0,3)o : | : | : | : - - + - - - - - - - - - - - + - - | 27
Its distance from $\displaystyle A(0,3)$ is: .$\displaystyle PA \:=\:\sqrt{x^2 + (y-3)^2}$
Its distance from the line $\displaystyle x = 27$ is: .$\displaystyle PB \:=\:27-x$
We are told that $\displaystyle PA$ is one-third of $\displaystyle PB.$
So we have the equation: .$\displaystyle \sqrt{(x^2 + (y-3)^2} \:=\:\frac{1}{2}(27-x)$
Now simplify the equation and identify the graph.