# Thread: Geometry of the Parabola

1. ## Geometry of the Parabola

Hey every1 hehe thnx in advance

P is a variable point on the parabola x^2=4y. The normal at P meets the parabola again at Q.The tangents at P and Q meet at T. S is the focus and QS=2PS

Prove that angle PSQ is right angle

2. ## Re: Geometry of the Parabola

Originally Posted by JustRandomThings
Hey every1 hehe thnx in advance

P is a variable point on the parabola x^2=4y. The normal at P meets the parabola again at Q.The tangents at P and Q meet at T. S is the focus and QS=2PS

Prove that angle PSQ is right angle
$\noindent \frac{dy}{dx} = \frac{x}{2}. If P\left(2p, p^2\right) then the normal at P has gradient -\frac{1}{p}. So the equation of the normal at P is y - p^2 = -\frac{1}{p}\left(x-2p\right). ie. y = -\frac{1}{p}x + 2 + p^2. To find Q solve this simultaneously with x^2 = 4y to get \frac{x^2}{4} = -\frac{1}{p}x+p^2+2. Rearranging, x^2 + \frac{4}{p}x - 4\left(p^2+2\right) = 0. Being a quadratic there are two solutions but one of which we already know (ie. x = 2p) so using the sum of roots, we find that Q\left(\frac{-2\left(p^2+2\right)}{p},\frac{\left(p^2+2\right)^2 }{p^2}\right).$

$\noindent All points on the parabola are equidistant from the focal point and the directrix, and since the directrix is y = -1 then the distances QS and PS are just one more than their y-coordinates. ie. PS =p^2 + 1 and QS = \frac{\left(p^2+2\right)^2}{p^2} + 1. Since QS = 2PS then \frac{\left(p^2+2\right)^2}{p^2} + 1 = 2\left(p^2 + 1\right). Simplifying, \left(p^2+2\right)^2 - 2p^2\left(p^2+1\right) + p^2 = 0. Then expand, simplify and factorise to get (p^2+1)(p-2)(p+2) = 0 which means p = \pm 2 since p \in \mathbb{R}. We only take p = 2 because the parabola is symmetrical. Using p we can find P and Q and then prove that m_{PS} \cdot m_{QS} = -1.$

3. ## Re: Geometry of the Parabola

Thanks a lot

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