# Finding tangent angle

Finding the tangent angle between the negative $$x$$-axis and the parabola $$y=-ax^2+bx$$($$a,b>0$$) at $$(x_0,y_0)$$ : I am trying to find the tangent angle with negative $$x$$ axis for a parabolic curve. I assume the equation of tangent line will be $$y=mx+C$$ and the equation for the parabola $$y=-a{x^2}+bx$$ So,$$-a{x^2}+bx-mx-C=0$$ Again $${y_0}=m{x_0}+C$$ We obtain $$a{x^2}+(m-b)x+{y_0}-m{x_0}=0$$ So if this tangent line is to be the desired tangent, then this $$x$$ has to be unique. That is, $$(m-b)^2-4(y_0-mx_0)a=0$$ Then $$m$$ has two values. So it seems to me complicated what $$m$$ I have to choose.Now what can I do?

• Can i say $\frac{dy}{dx}=\tan(π-\theta)$? – Raihan Amin Feb 13 at 9:38
• I suppose you are looking for the complementary angle so yes $\frac{dy}{dx} = \tan{(\pi - \theta)}$ is correct – Ahlfkushevich Feb 13 at 9:57
If you take into account that $$(x_0,y_0)$$ is on the parabola, the last equation in $$m$$ can be rewritten in the following way: \begin{align} (m-b)^2-4&\bigl(-ax_0^2+(b-m)x_0\bigr)a=m^2-2mb+b^2+4a^2x_0^2-4abx_0+4ax_0m \\ &=m^2+2(2ax_0-b)m+\underbrace{b^2+4a^2x_0^2-4abx_0}_{\textstyle(2ax_0-b)^2}\\[-1ex] &=\bigl(m+(2ax_0-b)\bigr)^2, \end{align} so that the last equation has a double root (in $$m$$), and you don't have to choose.