Geometry of the Parabola (1 Viewer)

JustRandomThings · Jul 12, 2017

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

1729 · Jul 13, 2017

JustRandomThings said:
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 $m_{PS} \cdot m_{QS} = -1$.$

JustRandomThings · Jul 13, 2017

Thanks a lot

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Geometry of the Parabola (1 Viewer)

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