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This was a homework assignment I had to do a while ago:

The curve K is given by the set of points $(x,y) \in \mathbb{R}^2$ such that $9x + 27y - \frac{10}{81}(x+y)^3 = 0 $. There is also a straight line $l$ tangent to $K$ which goes to $(0,0)$.

The assignment was: there are two points on $K$ such that the lines tangent to $K$ there intersect with $l$ perpendicularly. Find the coordinates of these points.

By means of implicit differentiation I was able to find $K' : 9 + 27y' - \frac{30}{81}(x+y)^2(1+y') = 0 $. Now we replace $y$ by $f(x)$ and $y'$ by $f'(x)$. Putting $(0,0)$ in $K'$ gives us: $ 9 + 27 f'(0) = 0 \implies f'(0) = - \frac{1}{3} x $. Now, $l : y = - \frac{1}{3}x + b $ We know that $l$ goes through $(0,0)$, so $0 = - \frac{1}{3} \cdot 0 + b \implies b=0$. So we find $l: y = - \frac{1}{3}x $ .

Furthermore, I also know that if we want to find the slope of the straight line $r$ that intersects $l$ perpendicularly, we have $slope_{l} \cdot slope_{r} = -1 \implies slope_{r} = \frac{-1}{- \frac{1}{3} } = 3 $.

But now what? How do I find the coordinates of the points I was assigned to find? I don't know where to plug in the $3$, if I even have to do so.

Thanks,

Max

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1 Answer

up vote 1 down vote accepted

Hints:

For your tangent(s) $r$, you know the slope is 3 so $y'=3$.

That should be enough to let you find $x+y$ for where $r$ touches $K$, and so $y$, and thus $x$

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Thank you! (text) –  Max Muller Nov 8 '11 at 16:16
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