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
