2
$\begingroup$

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 !

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.