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Problem
The angle between the tangent lines from the point $A(0,-1)$ to parabola defined as $y=x^2-ax+3$ is $135^{\circ}$. Then what could be the value of $a$?

My attempt
First I found that if $tan(\alpha)$ is $m$, i.e. slope of the first line, then $tan(\alpha +135^{\circ})$ must be $\frac{m-1}{m+1}$ since $tan(\alpha +\beta)=\frac{tan(\alpha)+tan(\beta)}{1-tan(\alpha)tan(\beta)}$. Say, the first line is tangent to the parabola at $x_1$, then the slope will be $2x_1-a$ (taking the derivative at $x_1$). Continuing with the similar variable naming, slope of the second tangent line will be $2x_2-a$.

Other facts:
$m=\frac{y_1-(-1)}{x_1-0}$ where $y_1=x_1^2-ax_1+3$ and $y_2=x_2^2-ax_2+3$
I proceed from here with $\frac{m-1}{m+1}=\frac{y_1-x_1+1}{y_1+x_1+1}=\frac{y_2-(-1)}{x_2}$. Then putting the formulas for $y_1$ and $y_2$ gives me another equality in terms of $x_1$ and $x_2$.

Using all these, I can't seem to extract $a$ and canceling other terms or relating above equations to slope found from derivative.

How to solve this problem, what should be done?

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HINT...A line passing through $(0,-1)$ with gradient $m$ is $$y=mx-1$$

If you solve this simultaneously with the curve $y=x^2-ax+3$ the resulting quadratic in $x$ must have double roots, and therefore zero discriminant, and this will give you two possible values of $m$ in terms of $a$.

You can then use the angle between the lines formula $$\tan\theta=\left|\frac{m_1-m_2}{1+m_1m_2}\right|$$ to find the four possible values of $a$

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  • $\begingroup$ I found $x^2-(m+a)x+4=0$ then $(m+a)^2=16$ since the discriminant is zero. Similarly, for the other line we have $x^2-(\frac{m-1}{m+1}+a)x+4=0)$ then equating both discriminants gave me $(m-1)/(m+1)+a=m+a$ or $(1-m)/(m+1)+a=m+a$. First one cannot happen because it gives a complex value for $m$. Using second one I got $m^2+2m-1=0$ and the roots are $-1-\sqrt{2}$ and $-1+\sqrt{2}$. Using $(m+a)^2=16$ gives me 4 different $a$ values, specifically, $-3+\sqrt{2}$, $-3-\sqrt{2}$, $5+\sqrt{2}$ and $5-\sqrt{2}$. Is it correct? $\endgroup$ – Motun Apr 10 '16 at 11:19
  • $\begingroup$ @Motun see my edited answer. $\endgroup$ – David Quinn Apr 10 '16 at 12:38
  • $\begingroup$ $tan(135^{\circ})=-1$, how can this be equal to absolute value of something? $\endgroup$ – Motun Apr 10 '16 at 12:46
  • $\begingroup$ Set the expression without the modulus sign equal to $\pm1$ since the angle between the tangents can be regarded as either 135 or 45 $\endgroup$ – David Quinn Apr 10 '16 at 12:49
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$$\dfrac{dy}{dx}=2x-a$$

So, the gradient$(m)$ at $(t,t^2-at+3)$ will be $2t-a\iff2t=m+a$

So, the equation of tangent at $(t,t^2-at+3)$ will be $$\dfrac{y-t^2-at+3}{x-t}=2t-a$$

It needs to pass through $(0,-1)\implies$ $$\dfrac{-1-t^2-at+3}{0-t}=2t-a\iff t^2-2at+2=0$$

Replacing $2t$ with $m+a$ $$m^2-2am+8-3a^2=0$$ which is a Quadratic equation in $m$

So, $m_1+m_2=2a, m_1m_2=8-3a^2$

$(m_1-m_2)^2=(m_1+m_2)^2-4m_1m_2=16a^2-32$

So, we need $16a^2-32\ge0\iff a^2\ge2$

WLOG $m_1>m_2\implies m_1-m_2=+\sqrt{16a^2-32}$

$$-1=\tan135^\circ=\dfrac{m_1-m_2}{1+m_1m_2}=\dfrac{\sqrt{16a^2-32}}{9-3a^2}$$

Clearly, $9-3a^2<0\iff a^2>3$

Square both sides and solve for $a^2$

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