# Parabola question

• Feb 6th 2014, 02:52 AM
Deci
Parabola question
So, okay I don't even know what the question means. I'm so confused now..

Consider the parabola $\displaystyle 4ay = x^2$, where $\displaystyle a > 0$ and suppose the tangents at $\displaystyle P(ap, ap^2)$ and $\displaystyle Q (2aq, aq^2)$ intersect at point T.

Let $\displaystyle S (0, a)$ be the focus of the parabola.

1) Find the coordinates of T. (assume that the tangents at P is $\displaystyle y = px - ap^2$).

2) Show that $\displaystyle SP = a(p^2 +1)$. Suppose P and Q move on the parabola in such a way that $\displaystyle SP + SQ = 4a$

3) Show that T is constrained to move on a parabola
• Feb 6th 2014, 03:30 AM
SlipEternal
Re: Parabola question
Have you done any work so far? Try typing up as much as you can.
• Feb 6th 2014, 07:55 AM
Deci
Re: Parabola question
O, o I solve (1) yay!

What I did is tangent P is equal to $\displaystyle y = px - ap^2$
while tangent Q is equal to $\displaystyle y = qx - aq^2$

$\displaystyle px - ap^2 = qx - aq^2$

$\displaystyle x = (p+q)a$

$\displaystyle y = p((p+q)a) - ap^2 = apq$

Im so happy right now..!

(2) is like this i guess..

$\displaystyle SP + SQ = 4a$

$\displaystyle SP + (a(q^2 - 1) = \frac{x^2}{y}$

$\displaystyle SP + (a(q^2 - 1) = \frac{ap+aq}{apq}$

$\displaystyle SP = \frac{1}{q} + \frac{1}{a}- (aq^2 - aq)$
• Feb 7th 2014, 01:08 PM
SlipEternal
Re: Parabola question
For (1), you have it correct. For (2), I am not sure what you are doing. You are supposed to show that $\displaystyle SP = a(p^2+1)$. You can do that by using the distance formula. LaTeX is not working, so plugging in the information to the distance formula:

$$SP = \sqrt{ (2ap - 0)^2 + (ap^2 - a)^2 }$$

Simplifying, you will get $\displaystyle SP = a(p^2+1)$ as you wanted. Similarly, $$SQ = a(q^2+1)$$.

Then, for (3), you suppose that $$SP+SQ=4a$$, then show that the point T will move along a parabola.

So, you get $$SP+SQ = a(p^2+1) + a(q^2+1) = a(p^2+q^2+2) = 4a$$. Hence, $$p^2 + q^2 = 2$$.

From the formula for T you found in part (1), you know $$x = a(p+q)$$. Squaring both sides, you get

$$x^2 = a^2(p+q)^2 = a^2(p^2+2pq+q^2) = a^2(p^2+q^2) + 2a(apq) = 2a^2 + 2ay$$

Solving for y, you get: $$y = \dfrac{1}{2a}x^2 - a$$ which is the definition for a parabola as desired.
• Feb 8th 2014, 03:04 AM
Deci
Re: Parabola question
Thanks SlipEternal
But im kind of confused in the distance formula,

SP = \sqrt{ (2ap - 0)^2 + (ap^2 - a)^2 }

But why do you use 2ap in (2ap - 0)^2 instead of ap ? P's x coordinate is ap right?
• Feb 8th 2014, 04:43 AM
SlipEternal
Re: Parabola question
Quote:

Originally Posted by Deci
Thanks SlipEternal
But im kind of confused in the distance formula,

SP = \sqrt{ (2ap - 0)^2 + (ap^2 - a)^2 }

But why do you use 2ap in (2ap - 0)^2 instead of ap ? P's x coordinate is ap right?

No, it is 2ap. You had a typo. The point (ap,ap^2) is not a point on the parabola. Here is why:

4ay = x^2 is the formula for the parabola. Plug in ap for x and ap^2 for y: 4a(ap^2) = 4a^2p^2 does not equal a^2p^2. On the other hand, if you change x to 2ap, on the right hand side you get 4a^2p^2, which is the same as the left hand side.