parabola

[Dr.Peterson]

However I seem to be having problems solving this. The points we are looking for have coordinates (x,y/2), put that in the y^2=2px and you get

y^2/4=2px
y^2=8px
obviously not correct

Essentially, you are compressing the graph vertically by a factor of 1/2. That is, as you say, given a point (x,y) on the original graph, (x,y/2) will be on the new graph. So if (x,y) is on the new graph (whose equation we want to write), then (x,2y) must be on the original graph.

Since the equation of the given graph is y^2 = 2px, we have to replace y with 2y, and we get (2y)^2 = 2px, so 4y^2 = 2px and y^2 = px/2.

This sort of substitution can be easy to get wrong!

Another approach to this, more like the usual way we solve locus problems, is to use a parameter. For any value of t, the point (t^2/(2p), t) is on the parabola (that is, I am taking y = t), so the corresponding point on the locus is (t^2/(2p), t/2). That is,

x = t^2/(2p)​

y = t/2​

To find the equation of the locus, we want to eliminate t. Solving the second equation for t, t = 2y; substituting in the first equation,

x =(2y)^2/(2p)​

2px = 4y^2​

y^2 = px/2​

 

[Loki123]

Essentially, you are compressing the graph vertically by a factor of 1/2. That is, as you say, given a point (x,y) on the original graph, (x,y/2) will be on the new graph. So if (x,y) is on the new graph (whose equation we want to write), then (x,2y) must be on the original graph.

Since the equation of the given graph is y^2 = 2px, we have to replace y with 2y, and we get (2y)^2 = 2px, so 4y^2 = 2px and y^2 = px/2.

This sort of substitution can be easy to get wrong!

Another approach to this, more like the usual way we solve locus problems, is to use a parameter. For any value of t, the point (t^2/(2p), t) is on the parabola (that is, I am taking y = t), so the corresponding point on the locus is (t^2/(2p), t/2). That is,

x = t^2/(2p)​

y = t/2​

To find the equation of the locus, we want to eliminate t. Solving the second equation for t, t = 2y; substituting in the first equation,

x =(2y)^2/(2p)​

2px = 4y^2​

y^2 = px/2​

while i do understand this, i am not sure why I can't get correct answer with (x,y/2)

 

[Dr.Peterson]

while i do understand this, i am not sure why I can't get correct answer with (x,y/2)

Because you're going in the wrong direction.

Read what I said very carefully to see why I used (x,2y) instead. Or, try to explain just as carefully why what you did should be right. Sometimes when you try to explain your thinking, you can see where it is wrong.

 

[Loki123]

Because you're going in the wrong direction.

Read what I said very carefully to see why I used (x,2y) instead. Or, try to explain just as carefully why what you did should be right. Sometimes when you try to explain your thinking, you can see where it is wrong.

Okay, thank you, I think I get it.