2 tangents to the parabola $y^2=4ax$ meet at an angle of $45^\circ$. Prove that the locus of their point of intersection is $y^2-4ax=(x+a)^2$.
$$$$
I got completely stuck with this question. All I could think of was that the tangents would be of the form $$y=m_1x-am_1^3-2am_1$$ and $$y=m_2x-am_2^3-2am_2$$ where $m_1$ and $m_2$ are the slopes of the respective tangents. I also know that $\tan 45=|\dfrac{m_1-m_2}{1-m_1m_2}|$ $$$$
Using this, I tried to work out the value of $m_2$ in terms of $m_1$. However I got stuck because there were 2 possible values of $m_2$ that I got from the $\tan 45$ equation, and I couldn't decide which value of $m_2$ to use.
$$$$Lastly, I know the result that if there are 2 tangents at $t_1$ and $t_2$, their point of intersection is at $(at_1t_2, a(t_1+t_2)$
However, I cannot understand how to apply these results, or to use some other method. I would be truly grateful if someone could please help me with this problem. Many thanks!
Best Answer
Corresponding to parameter value $t$, the gradient of the tangent is $\frac 1t$.
Therefore, from the angle formula, you have $$\left|\frac{\frac{1}{t_2}-\frac{1}{t_1}}{1+\frac{1}{t_1t_2}}\right|=\left|\frac{t_1-t_2}{1+t_1t_2}\right|=1$$
You also have $\frac xa=t_1t_2$ and $\frac ya=t_1+t_2$
Writing the first equation as $$|t_1-t_2|=|1+\frac xa|$$ and squaring both sides, we get $$\left(1+\frac xa\right)^2=(t_1-t_2)^2=(t_1+t_2)^2-4t_1t_2=\left(\frac ya\right)^2-4\frac xa$$
Hence the result follows immediately