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

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?

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 $$P,(1,0)$$

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

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

• I don't understand, how did you arrive to that? – davidllerenav Apr 12 '19 at 4:38
• @david, Please pinpoint your confusion – lab bhattacharjee Apr 12 '19 at 4:41
• I don't understand any point is $P (t^2,2t)$. – davidllerenav Apr 12 '19 at 4:52
• @davis, See nabla.hr/PC-Parabola3.htm – lab bhattacharjee Apr 12 '19 at 5:02
• I'll check it out as soon as possible. But how can I solve this usiing Lagrange multipliers, since this is the way I'm supposed to solve the problem. – davidllerenav Apr 12 '19 at 5:16

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

• Ok, so I need to use $y=0$, not $\lambda=1$? – davidllerenav Apr 12 '19 at 4:54
• Yes, that is right. – Prajwal Kansakar Apr 12 '19 at 5:01
• Ok, so with $y=0$ I get $x=0$ by replacing $y$ on $y^2=4x$, right? – davidllerenav Apr 12 '19 at 5:07