parabola question. (2 Viewers)

[barbernator]

the normal to the parabola x^2=4ay with equation x+py=2ap+ap^3

how many values of p allow the normal to pass through the focus.

now the intuitive answer would be 3, yet upon substituting in (0,a) into the equation and solving for p, it results in 2 imaginary solutions and only 1 real solution. why is this?

 

[seanieg89]

Why would the intuitive answer be three? Cubics don't necessarily have three real roots.

As soon as you start to get close to the vertex of the parabola, you reach a bounding curve. When you pass this curve you can only draw one normal to the parabola from your point. You can actually calculate the equation of this curve without too much effort by using the cubic discriminant formula.

 

[barbernator]

Why would the intuitive answer be three? Cubics don't necessarily have three real roots.

As soon as you start to get close to the vertex of the parabola, you reach a bounding curve. When you pass this curve you can only draw one normal to the parabola from your point. You can actually calculate the equation of this curve without too much effort by using the cubic discriminant formula.

Well what i thought is that if we have a variable normal to move around, as you move just to the left of the vertex, the normal will cross to the right of the focus. As you slide it up the left branch of the parabola, it will eventually most to the right of the focus. At some point there must be a cross over where it passes through the focus? and this happens on both sides so 3 solutions?

 

[RealiseNothing]

Well what i thought is that if we have a variable normal to move around, as you move just to the left of the vertex, the normal will cross to the right of the focus. As you slide it up the left branch of the parabola, it will eventually most to the right of the focus. At some point there must be a cross over where it passes through the focus? and this happens on both sides so 3 solutions?
View attachment 26228

Are you sure that happens though? I would think that moving it to the left would put the line just on the left of the focus.

 

[barbernator]

Are you sure that happens though? I would think that moving it to the left would put the line just on the left of the focus.

yeh haha thats probably right, death by diagram hey sean

 

[seanieg89]

Well what i thought is that if we have a variable normal to move around, as you move just to the left of the vertex, the normal will cross to the right of the focus. As you slide it up the left branch of the parabola, it will eventually most to the right of the focus. At some point there must be a cross over where it passes through the focus? and this happens on both sides so 3 solutions?
View attachment 26228

The problem is your assumption that the normal will "cross" to the right side of the focus. This does not happen, nor is the diagram any guarantee that it should. The same sort of "argument by pictures" is what can lead to fallacies like the equilateral triangle thing.

 

[seanieg89]

yeh haha thats probably right, death by diagram hey sean

Haha yep. This sort of thing is why we need rigour in mathematics and can't just believe things because they look true.

 

[Carrotsticks]

I believe there was a Sydney Boys MX2 paper Q8 based on a very similar concept.