# Geometry problem with parabola and two tangents which intersect

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.

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 $$x_A^2 - y_A^2 - 12y_A - 4 = 0$$

• You are an absolute genius for me – Aleksandra Asanin Aug 25 '19 at 21:04
• No, it's not so. I like to help someone if I can. – Luca Goldoni Ph.D. Aug 26 '19 at 5:55

$$x^2=8y$$

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:
$$c_1 = y_1-m_1x_1 = \dfrac{{x_1}^2}{8} - m_1x_1 = \dfrac{(4m_1)^2}{8} - m_1(4m_1) = -2{m_1}^2$$

• 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 – Aleksandra Asanin Aug 25 '19 at 18:51
• 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? – AgentS Aug 25 '19 at 18:54
• Was a typo.. :) – Aleksandra Asanin Aug 25 '19 at 18:56
• cool:) for #2 solve the system of equations : $$y=m_1x -2m_1^2\\y=m_2x-2m_2^2$$ – AgentS Aug 25 '19 at 19:00
• 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. – Aleksandra Asanin Aug 25 '19 at 19:02

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: $$c_1=-m_1^2$$.