Find the differential equation of all tangent lines of parabola $y^2=4x$

Question

My professor said that it's $x(y')^2-yy'+1=0$ but how? I drew it and I think it open to the right $90^\circ$ but I can find the solution to differentiate

Answer 1

We have $$2y\frac{dy}{dx}=4$$ and so $$\frac{dy}{dx}=\frac2y\ ,\quad y\ne0\ .$$ The tangent line when $y=b\ne0$ is $$y-b=\frac2b\Bigl(x-\frac{b^2}4\Bigr)$$ which simplifies to $$y=\frac2bx+\frac b2\ .\tag1$$ Hence the line satisfies $$y'=\frac2b\ ;\tag2$$ combining with the previous equation, $$y=y'x+\frac1{y'}$$ which simplifies to the equation you were given.

Answer 2

Differentiating implicitly both sides to get: $2yy' = 4\Rightarrow 4y^2(y')^2=16\Rightarrow (y')^2 = \dfrac{4}{y^2}=\dfrac{1}{x}\Rightarrow x(y')^2 = 1$. Also: $yy'=\dfrac{4}{2}=2$,and since $1-2+1=0$, substituting we have: $x(y')^2-yy'+1=0$.