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!


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 think that this is not correct.

Let $(s,t)$ be a point on the parabola. Then, since $m=\frac{2a}{t}\Rightarrow t=\frac{2a}{m}$ with $t^2=4as$, $$y-t=\frac{2a}{t}(x-s)\iff y-\frac{2a}{m}=m\left(x-\frac{1}{4a}\left(\frac{2a}{m}\right)^2\right)$$So, $$y=m_1x+\frac{a}{m_1},\quad y=m_2x+\frac{a}{m_2}$$

I also know that $\tan 45=|\dfrac{m_1-m_2}{1-m_1m_2}|$

Note that we have $$\tan(45^\circ)=\left|\frac{m_1-m_2}{1\color{red}{+}m_1m_2}\right|\tag1$$

Since the intersection point $(X,Y)$ is $\left(\frac{a}{m_1m_2},\frac{m_1+m_2}{m_1m_2}a\right)$, we have $$m_1m_2=\frac{a}{X},\quad m_1+m_2=\frac YX.$$

Hence, from $(1)$ $$(1+m_1m_2)^2=(m_1-m_2)^2=(m_1+m_2)^2-4m_1m_2$$ $$\left(1+\frac aX\right)^2=\left(\frac YX\right)^2-4\cdot\frac aX,$$ i.e. $$Y^2-4aX=(X+a)^2$$

  • $\begingroup$ I'm really sorry. I got confused with the equations of tangents and normals. Really sorry once again. Thanks very much for your help and patience. $\endgroup$ – Better World Jan 7 '16 at 8:22
  • $\begingroup$ Sir, could you please explain why you've equated with the expression inside the modulus with $1$ instead of $\pm 1$? $\endgroup$ – Better World Jan 7 '16 at 8:27
  • $\begingroup$ @BetterWorld I think he has equated the square of that expression to $1$ which is perfectly fine. $\endgroup$ – najayaz Jan 7 '16 at 8:28
  • $\begingroup$ But Sir, if we have say line $l_1$ with slope $m_1$ which is known, then there are 2 possible lines $l_2$ and $l_3$ with slope $m_2$ and $m_3$ respectively which could make a $45^\circ$ angle with $l_1$. However, from $\tan(45^\circ)=\frac{m_1-m_2}{1+m_1m_2}$ we woulcdd get only one value of $m$ which could be the value of $m_2$ or $m_3$. This, we would get the slope of only one of the 2 possible lines. Isn't that why we use $\tan(45^\circ)=\tan(45^\circ)=\left|\frac{m_1-m_2}{1+m_1m_2}\right|$, to get both the slopes of both possible lines $l_2$ and $l_3$? $\endgroup$ – Better World Jan 7 '16 at 10:17
  • $\begingroup$ @BetterWorld: Sorry, forget my last comment (which I already deleted). We have $(1)$ in my answer, which is $1=\left|\frac{m_1-m_2}{1+m_1m_2}\right|$. So, we have $|1+m_1m_2|=|m_1-m_2|$. Therefore, we have $(1+m_1m_2)^2=(m_1-m_2)^2$. Any confusion about these steps? $\endgroup$ – mathlove Jan 7 '16 at 10:25

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

  • $\begingroup$ Sir, could you please explain why $\dfrac{t_1-t_2}{1+t_1t_2}=1$ instead of $\dfrac{t_1-t_2}{1+t_1t_2}=\pm1$? $\endgroup$ – Better World Jan 7 '16 at 8:23
  • $\begingroup$ sorry, I have edited my answer! $\endgroup$ – David Quinn Jan 7 '16 at 8:36

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