Geometry problem with parabola and two tangents which intersect

Question

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.

Answer 1

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

Answer 2

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 $(x,y)$.

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

Answer 3

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: $c_1=-m_1^2$.