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If the normal at $P(ap^2,2ap)$ to the parabola $y^2=4ax$ meets the curve again at $Q(aq^2,2aq)$, show that $p^2+pq+2=0$. Show that the equation of the locus of the point of intersection of the tangents at $P$ and $Q$ to the parabola is $y^2(x+2a)+4a^3=0$.

Blue
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黃偉恆
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  • Is the point P and Q identical? – Dr. Sonnhard Graubner Aug 23 '19 at 18:46
  • no, typing wrongly.I'm edited the question . – 黃偉恆 Aug 23 '19 at 19:10
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  • The first part of your question is a duplicate of https://math.stackexchange.com/questions/3236502/the-normal-at-tat2-2at-of-parabola-y2-4ax-meets-the-parabola-again-at-s?rq=1, which you can find in the handy list of related questions at right. – amd Aug 23 '19 at 20:55

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The equation of the parabola is $x=\frac{y^2}{4a}$. I know that the angular coefficent of the tangent line to the parabolais $m=2ay_0+b=2\cdot \frac{1}{4a}y_0=\frac{y_0}{2a}$ where $y_0$ is the $y$ coordinate of the tangency point and $a,b$ are coefficents of $x=ay^2+by+c$. The tangent line so is: $x-ap^2=\frac{1}{2a}(y-2ap)$ so $x+ap^2=py$. The normal line has angular coefficent $m_1=-\frac{2a}{y_0}$. The normal has equation: $x-x_0=-\frac{2a}{y_0}(y-y_0)$. Substituing the values of $x_0$ and $y_0$, I obtain: $x-ap^2=-\frac{1}{q}(y-2ap)$. Now, the solutions of the system: $$\left\{\begin{matrix} x-ap^2=-\frac{1}{q}(y-2ap) \\ x=\frac{y^2}{4a} \end{matrix}\right.$$ are: $x=aq^2$ and $y=2aq$. Substituing another time, I have: $$a(q-p)(q+p)=-\frac{1}{q}\cdot 2a(q-p)$$ and so: $$q^2+pq+2=0$$

Matteo
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