How to find the shortest distance from $(1,0)$ to $y^2=4x$?

Question

I need help with this problem:

Find by the method of Lagrange multiplier the shortest distance from the point $(1,0)$ to the parabola $y^2=4x$. Check your answer by a method of substitution.

Answer: $1$.

I first selected $f(x,y)=(x-1)^2+y^2$ as the function that I need to minimize, since it is the shortest distance formula. Then I think that $g(x,y)=y^2-4x$ is the constraint. So, by using the method of Lagrange multipliers: $$(grad \ g)(x,y)=\lambda(grad \ f)(x,y)$$ $$(-4,2y)=\lambda(2(x-1),2y) $$ $$\Rightarrow -4=2x\lambda-2\lambda$$ $$\Rightarrow 2y=2y\lambda$$ thus $\lambda=1$ and by replacing this on the othe equation, I get $x=-1$. What is $y$ equal to? I tried to find it by replacing x int the parabola equation, but I ended up with $y=\sqrt{-4}$, what am I doing wrong?

Answer 1

It is much easier to use the following method

WLOG any point on $y^2=4x$ is $P(t^2,2t)$

If $d$ is the distance between $P,(1,0)$

$$d^2=(t^2-1)^2+(2t-0)^2=(t^2+1)^2$$

$\implies d=t^2+1\ge1$

Answer 2

$2y=2y\lambda$ implies either $\lambda=1$ OR $y=0$. Since $\lambda=1$ is not valid as you showed so $y=0$.