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Given a curve $P: x^2 = 8y$. Two tangents to $P$ are $L_1: y=m_1x+c_1$ and $L_2: y=m_2x+c_2$, and they are intersecting at a point $A$.

Problem 1. Express $c_1$ in terms of $m_1$

  • Solution: $c_1 = - 2m_1 ^2$ $\leftarrow$ I don't know how $x$ is eliminated here.

Problem 2. Show that coordinates of $A$ are $(2(m_1+m_2), 2m_1m_2)$

  • This reminds me of the angle between lines formula... but the angle is not given at this point, hence I don't know where to start.

Problem 3. If the angle between $L_1$ and $L_2$ is $\dfrac{\pi}{4}$, find the equation of the locus of $A$.

  • Solution: $x^2-y^2-12y-4=0$ $\leftarrow$ Have no idea how to get here.

Anyone, please help, I am freaking out.

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First question:

  1. The line and the parabola are tangent therefore the system $$ \left\{ \begin{gathered} x^2 = 8x \hfill \\ y = m_1 x + c_1 \hfill \\ \end{gathered} \right. $$ must have two coincident solutions. The system get to you the following equation $$ x^2 - 8m_1 x - 8c_1 = 0 $$

which has two coincident solution if and only if discriminant is zero. It means that $$ 64m_1 ^2 + 32c_1 = 0 $$ thus $$ c_1 = - 2m_1 ^2 $$ In the same way you get that $$ c_2 = - 2m_2 ^2 $$ hence your lines are $$ \begin{gathered} L_1 :y = m_1 x - 2m_1 ^2 \hfill \\ L_2 :y = m_2 x - 2m_2 ^2 \hfill \\ \end{gathered} $$

Second question

The point A is the common point of the two lines thus it is the solution of the system $$ \left\{ \begin{gathered} y = m_1 x - 2m_1 ^2 \hfill \\ y = m_2 x - 2m_2 ^2 \hfill \\ \end{gathered} \right. $$ You have $$ m_1 x - 2m_1 ^2 = m_2 x - 2m_2 ^2 $$ which means $$ x\left( {m_1 - m_2 } \right) = 2\left( {m_1 ^2 - m_2 ^2 } \right) = 2\left( {m_1 - m_2 } \right)\left( {m_1 + m_2 } \right) $$ Since the lines are not parallel you have that $$m_1 \neq m_2$$ and therefore you have $$ x = 2\left( {m_1 + m_2 } \right) $$ By substitution you have $$ y = 2m_1 \left( {m_1 + m_2 } \right) - 2m_1 ^2 = 2m_1 m_2 $$ These are therefore the coordinates of A.

Third question

You have that $$ \left| {\frac{{m_1 - m_2 }} {{1 + m_1 m_2 }}} \right| = \tan \theta $$ In your case $$ \left| {\frac{{m_1 - m_2 }} {{1 + m_1 m_2 }}} \right| = 1 $$ You can drop the absolute value by squaring: $$ \left( {\frac{{m_1 - m_2 }} {{1 + m_1 m_2 }}} \right)^2 = 1 $$ It follows that $$ \left( {m_1 - m_2 } \right)^2 = \left( {1 + m_1 m_2 } \right)^2 $$ and so $$ m_1 ^2 + m_2 ^2 - 2m_1 m_2 = 1 + 2m_1 m_2 + \left( {m_1 m_2 } \right)^2 * $$ Since $$ m_1 ^2 + m_2 ^2 = \left( {m_1 + m_2 } \right)^2 - 2m_1 m_2 $$ you can rewrite * as $$ \left( {m_1 + m_2 } \right)^2 - 6m_1 m_2 = 1 + \left( {m_1 m_2 } \right)^2 $$ Since, from question 2, you got that $$ \left\{ \begin{gathered} m_1 + m_2 = \frac{{x_A }} {2} \hfill \\ m_1 m_2 = \frac{{y_A }} {2} \hfill \\ \end{gathered} \right. $$ by substitution you get $$ x_A^2 - y_A^2 - 12y_A - 4 = 0 $$

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  • $\begingroup$ You are an absolute genius for me $\endgroup$ – Aleksandra Asanin Aug 25 '19 at 21:04
  • $\begingroup$ No, it's not so. I like to help someone if I can. $\endgroup$ – Luca Goldoni Ph.D. Aug 26 '19 at 5:55
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I'm going to explain #1 which may help you figure the rest on your own.

$x^2=8y$

Differentiate both sides and isolate $y'$: $$y' =\dfrac{x}{4}$$ Above is the slope of the tangent line to the parabola at $(x,y)$.


Next look at the first tangent line

$y = m_1x+c_1$
Say this tangent line touches the parabola at $(x_1,y_1)$, then the slope of the tangent line at this point is $m_1 = \dfrac{x_1}{4}$

Since the point $(x_1,y_1)$ is on both parabola and tangent line, it satisfies both:
$c_1 = y_1-m_1x_1 = \dfrac{{x_1}^2}{8} - m_1x_1 = \dfrac{(4m_1)^2}{8} - m_1(4m_1) = -2{m_1}^2$

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  • $\begingroup$ Helpful, but still don't know how to proceed... At least I am one step ahead. Didn't even know that I should use derivatives $\endgroup$ – Aleksandra Asanin Aug 25 '19 at 18:51
  • $\begingroup$ Hey @AleksandraAsanin your solution has $c_1=-{m_1}^2$, but as you can see I got $c_1=-2{m_1}^2$... are you sure there isn't a typo? $\endgroup$ – AgentS Aug 25 '19 at 18:54
  • $\begingroup$ Was a typo.. :) $\endgroup$ – Aleksandra Asanin Aug 25 '19 at 18:56
  • $\begingroup$ cool:) for #2 solve the system of equations : $$y=m_1x -2m_1^2\\y=m_2x-2m_2^2$$ $\endgroup$ – AgentS Aug 25 '19 at 19:00
  • $\begingroup$ yeees, would be pleased. I am not a geometry person at all, for me it would be easier to solve a triple integral than this. $\endgroup$ – Aleksandra Asanin Aug 25 '19 at 19:02
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The angular coefficent of the $L_1$ (knowing that the general coefficent $m=2ax_0+b$ for $ax^2+bx+c$) is: $m_1=\frac{1}{4}x_0$, so the line $L_1$: $y=\frac{1}{4}x_0x+c_1$. Now, the $\Delta$ of the following system has to be $0$: $$\left\{\begin{matrix} y=\frac{1}{4}x_0x+c_1 \\ y=\frac{x^2}{8} \end{matrix}\right.$$

Substituing and imposing $\Delta=0$, I have: $4x_0^2+32c_1=0$ so $c_1=-\frac{x_0^2}{8}$. Comparing $c$ and $m$, I have: $c_1=-m_1^2$.

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