Finding the point on a parabola closest to the origin

[leprofece]

Find the curve coordinates of the point nearest to P in the curve
2y² = 5(x+1) P(0,0)
Book answer (-1,0)

ok (x-0)²+ (y-0)² = D
D = (x)²+ (y)²
and (y)² = -x²

Now -x² = 5/2(x+1)

But derivating I don't get the answer or ansa

 

[shamieh]

Re: max and min 317

isn't the formula like

$$\sqrt{(x -0)^2 + (y- 0)^2} = D$$ ?

I think you forgot the square root . I'm not sure though but I think that cancels your squares in the end

 

[MarkFL]

leprofece, Please use thread titles that describe the problem.

You can in fact minimize the square of the distance for simplification of the computation:

$$f(x,y)=x^2+y^2$$

constrained by:

$$x=\frac{2}{5}y^2-1$$

So, substituting for $x$ into the objective function, we find:

$$f(y)=\left(\frac{2}{5}y^2-1 \right)^2+y^2$$

Now, differentiate with respect to $y$ and equate the result to zero to find the critical value.

 

[leprofece]

leprofece, Please use thread titles that describe the problem.

You can in fact minimize the square of the distance for simplification of the computation:

$$f(x,y)=x^2+y^2$$

constrained by:

$$x=\frac{2}{5}y^2-1$$

So, substituting for $x$ into the objective function, we find:

$$f(y)=\left(\frac{2}{5}y^2-1 \right)^2+y^2$$

Now, differentiate with respect to $y$ and equate the result to zero to find the critical value.

One Thing
Can I solve for y and constrain with x instead of y how you did?

 

[MarkFL]

One Thing
Can I solve for y and constrain with x instead of y how you did?

Yes, but then you run into a "problem." Try it and see...:D

It is still workable, but becomes a boundary problem like the circle problem you recently posted.