TITLE: Find the differential equation of all straight lines tangent to parabola $y=x^2$
QUESTION [0 upvotes]: Given a point P(a,b) on or outside the parabola $y=x^2$, how to find equation of straight line tangent to the parabola?

REPLY [1 votes]: Let $y=mx+c$  be tangent on $y=x^2\implies mx+C=x^2 \implies X^2-mx=C=$
For tangency $B^2=4AC \implies m^2=-4C \implies C=-m^2/4$
So $$y=mx-m^2/4~~~(1)$$ is one-parameter family of tangents to the given parabola.
Eliminating $m$ by differentiating (1) w.r.t. $x$, we get
$y'=m$ putting this in (1), we get first order non-linear differential equation as
$$y=xy'-(y')^2/4.$$ This is the required ODE which is of the type known as Clairaut.
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