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