1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Derivatives will help you $\endgroup$ – Ahlfkushevich Feb 13 at 9:22
  • $\begingroup$ Can i say $\frac{dy}{dx}=\tan(π-\theta)$? $\endgroup$ – Raihan Amin Feb 13 at 9:38
  • $\begingroup$ 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 $\endgroup$ – Ahlfkushevich Feb 13 at 9:49
  • $\begingroup$ I am just interested to the angle of tangent, at a given point of parabola, with the negative x axis. $\endgroup$ – Raihan Amin Feb 13 at 9:53
  • $\begingroup$ I suppose you are looking for the complementary angle so yes $\frac{dy}{dx} = \tan{(\pi - \theta)}$ is correct $\endgroup$ – Ahlfkushevich Feb 13 at 9:57
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.