4

The conic section k: $x^2+3y^2-2x+6y-8=0$. Find all points such that the normal vector of the conic section in these points is parallel to y-axis.

My approach: Find the gradient of the conic section:

$F(x,y) = x^2+3y^2-2x+6y$

$\nabla F(x,y) = <2x-2,6y+6>$

The y-axis vector is $(0,1)$. Find such constants that the gradient of the conic section is scalar multiple of y-axis:

$2x-2=0 \rightarrow x=1$

$6y+6=1 \rightarrow y=-5/6$

The point $[1,-5/6]$ however isn't the correct answer. If I consider the points separately, then for $x$ I get points $[1,1], [1,-3]$ which is a correct answer but I'm still left with $y=-5/6$ for which I get valid $x$'s but the points are not correct (in fact their are perpendicular to the y-axis and not parallel.

Where do I make mistake? Why am I getting the extra $y=-5/6$?

Owl City
  • 43
  • did you already see the plot and where is the direction $[1,-5/6]? if not then check http://www.wolframalpha.com/input/?i=+x%5E2%2B3y%5E2%E2%88%922x%2B6y%E2%88%928%3D0.+ – janmarqz Dec 24 '14 at 15:26

2 Answers2

1

You found $x$ correctly. However, as you said, the normal vector has to be a scalar multiple of $(0,1)$ which implies it has to be of the form $(0,n)$. This gives us:$$y=\frac{n}{6}-1$$If you substitute $x=1$ and $y=\frac{n}{6}-1$ into your original equation:$$x^2+3y^2-2x+6y-8=0$$you can solve this for $n$ and you should find:$$n=\pm12$$which gives you the correct values for $y$

Mufasa
  • 5,434
  • Oh I think I see it. I wasn't looking in fact for the scalar multiplier but for the points. I should've put the equations as: $i(2x-2)=0 \rightarrow xi=i \rightarrow x=1$ and $j(6y+6)=1 \rightarrow 6yj = 1-6j \rightarrow y=\frac{1}{6j}-1$. If I substitute this for $x$ and $y$ in the original equation I get $j=\pm 1/12$ which gives me $y=1$ and $y=-3$. Thanks. – Owl City Dec 24 '14 at 15:51
0

Hint: at the points $(1,1)$ and $(1,-3)$ the gradient direction is $(1,-5/6)$ for both. I hope the plot helps enter image description here

janmarqz
  • 10,538