In my Calculus I assigment, I'm stuck on the following :

Find $M_1=(x_1,y_1)$ on $y=5x+6$ and $M_2=(x_2,y_2)$ on $y=-(x-3)^2+4$ such that the square of the distance between $M_1$ and $M_2$ is minimal.

I'm fine with that, but they add the following : formulate the problem using a fourth degree polynomial implying only variables $x_1$ and $x_2$.

I did :

Be $D=(x_1-x_2)^2+((5x_1+6)-(-(x_2-3)^2+4))$ the distance we want to minimize, then $D=x_1^2+2x_2^2-2x_1x_2+5x_1-6x_2+11$ when expanded.

I don't understand where that "fourth degree polynomial" will appear?

Thanks !

  • $\begingroup$ Double check the expression for $D$, you need $(y_1 - y_2)^2$. $\endgroup$ – Macavity Nov 28 '13 at 5:11
  • $\begingroup$ @user111288: maybe you can translate both the line and the parabola, so that the line goes thru the origin. Then you can find the ortho projection of a general point in $y=x^2$ into the line. $\endgroup$ – user99680 Nov 28 '13 at 5:35

Answered in comment: the correct formula for squared distance, $$D=(x_1-x_2)^2+((5x_1+6)-(-(x_2-3)^2+4))^2$$ is indeed a fourth degree polynomial in $x_1,x_2$.

The straightforward approach is to equate $\partial D/\partial x_1$ to zero (this is a linear equation for $x_1$), and eliminate $x_1$ from $D$. The deal with $\partial D/\partial x_2=0$. Not pleasant, but maybe this is the purpose of the exercise.

A smarter approach is to realize that the tangent line to the parabola at $M_2$ must be parallel to the line $5x+6$. This gives $x_2$ right away.


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