# Find the equation of the following uniparametric family of curves:

Find the equation of the following uniparametric family of curves: All tangent lines to the parabola $$x^2 = \alpha y$$. Consider the value of $$\alpha$$ equal to the last digit of your DNI that is greater than or equal to $$2\, (\alpha ≥ 2 ).$$

For my case $$\alpha= 8$$.

Then I did the following: Any point on the parabola has to be of the form $$P=(t, \frac{t^2}{\alpha})$$, where $$t$$ is a parameter that can take any real value, and I assume that $$\alpha$$ is not zero.

Since the derivative of the function $$y=\frac{x^2}{\alpha}$$ is $$y'=\frac{2x}{\alpha}$$, the slope of the tangent line to the parabola at $$P$$ is $$\frac{2t}{\alpha}$$.

Therefore, the equation of the tangent line at $$P$$ to the parabola is: $$y -\frac{ t^2}{\alpha} =\frac{ 2t}{\alpha}(x-t).$$ That is to say: $$y- \frac{t^2}{8}= \frac{2t}{8}(x-t)$$. Guys, I don't know if this exercise is correct. Now I don't know if it's me who is interpreting wrong exercise since this is the class of setting up and solving ordinary differential equations and Differential equation of a family of curves

• Your solution is correct. Sep 18 at 21:37
• Thanks, very much. Could you review this demo I posted? Show that the curves of the second family are solutions of the DE Sep 18 at 22:41
• I don't see any demo. Sep 18 at 22:44
• math.stackexchange.com/posts/4770889/revisions Sep 18 at 22:44