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?
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Derivatives will help you – JoseSquare Feb 13 '19 at 09:22
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Can i say $\frac{dy}{dx}=\tan(π-\theta)$? – Raihan Amin Feb 13 '19 at 09:38
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What you are doing its just getting the intersection point between a general parabola and a general line, you are not using the condition that the line has to be tangent to the parabola – JoseSquare Feb 13 '19 at 09:49
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I am just interested to the angle of tangent, at a given point of parabola, with the negative x axis. – Raihan Amin Feb 13 '19 at 09:53
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I suppose you are looking for the complementary angle so yes $\frac{dy}{dx} = \tan{(\pi - \theta)}$ is correct – JoseSquare Feb 13 '19 at 09:57
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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.
Bernard
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