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.
Best Answer
First question:
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 $$ x_A^2 - y_A^2 - 12y_A - 4 = 0 $$