Parameter Question help. (1 Viewer)

[Tsylana]

Sounds simple enough...
Looks simple enough...
Creates enough of a headache.

The point p(2ap, ap^2) lies on the parabola x^2 = 4ay. The focus S is the point , (0,a) .The tangent at P meets the y axis at Q.

Prove PS = QS

xD. Help please? XD

 

[Tsylana]

lol... i tried that ><"

i ended up getting a huge bunch of crap.... that didn't equal xD...

 

[Aplus]

You prove that Q is the directrix. The definition of a parabola states that the distance of a point on the parabola is equidistant from both the focus and the directrix at any given point.

 

[tommykins]

Are you sure it's not PS = QS ?

I tried it out, doesn't work.

 

[12o9]

Are you sure it's not PS = QS ?

I tried it out, doesn't work.

yeah same here.

edit: nvm the question got edited x).

 

[Tsylana]

Already editted. >_>"

Lol @ Aplus... ><" makes me feel like an idiot rofl.

 

[Aplus]

Yeah, I tried it and panicked because I couldn't prove it. Good thing it wasn't the question.

 

[tommykins]

Yeah, I tried it and panicked because I couldn't prove it. Good thing it wasn't the question.

Your idea was incorrect as tangents of focal chords intersect at the directrix, it doesn't necesarily mean a tangent at a point cutting the y axis must be at the directrix ie (0,-a).

Also, PS = PD (D is directrix) but D is (x,-a) and it is the PERPINDICULAR distance from the point to the directrix.

 

[Tsylana]

*edit for the 3rd time. is just very confused now.*

 

[tommykins]

I got 0 ?

My first working was that PD (where D is the directrix) is
sqrt[(x-x)² + (ap²+a)²]
= sqrt[0 + (ap²+a)²]
= (ap²+a)

Now, the height of the triangle is simply x.

Therefore area of triangle = x.(ap²+a)/2

For the area underneath the parabola -
x² = 4ay
y = x²/4a

1/4a int x² dx x->0

1/4a [x³/3]x->0
= x³/12a

Area of Triangle/Area below curve = 6a²(p²+1)/x²

I'm abit iffy about that answer however, as there is a p variable in there.

If however, I find PD to be 4a (splitting the triangle into two right angled triangles, height of half of PD is 2a), the area of the triangle is 2a.x

2a.x/x³/12a -> 24a²/x²

 

[3unitz]

I got 0 ?
My first working was that PD (where D is the directrix) is
sqrt[(x-x)² + (ap²+a)²]
= sqrt[0 + (ap²+a)²]
= (ap²+a)

PS = PD = (x^2/4a + a)

 

[tommykins]

Silly me.

 

[Iruka]

That was a nice question, 3unitz. I am sure I will be able to put it to evil uses some day.

 

[tommykins]

Okay, I got 3/2 great question.

 

[HalcyonSky]

Okay, I got 3/2 great question.

^_^ good question ay i made it up all by myself

- 3unitz

 

[tommykins]

Haha honestly, I adore you, Slidey and Affinity so much. <3 math.

 

[Tsylana]

>_>"

Lol. I did this question for my mate a while ago and i handed him the solution but now and he lost it... & I can't remember how to do it.. could someone post up the solution... noting Q is at (0,-ap^2) for simplification.