# Intersection of normal to the curve

The line that is normal to the curve $x^2+3xy-4y^2=0$ at $(6,6)$ intersects the curve at what other point?

If I implicitly differentiate this curve, I will get the equation of the slope:

$$2x+3xy\prime+3y-8yy\prime=0 \implies y\prime=\frac{-2x-3y}{3x-8y}$$

If I evaluate this at the given points, I will get the tangent slope. The negative reciprocal of the tangent line's slope would the the normal line's slope.

How would I find other points which the normal intersects? Thanks.

• Find the equation of the normal, perhaps in form $y=ax+b$. Substitute $ax+b$ for $y$ in the equation of the curve, and solve for $x$. You will end up with a quadratic. We can then use the Quadratic Formula, though you can save some computation by using facts about the sum or product of the roots, since you know that one of them is $6$. Dec 11 '13 at 21:40

Since the curve is defined by the equation $$F(x,y)=x^2+3xy-4y^2=0,$$ the line that is normal to it at the point $p=(6,6)$ is given by $$N_p=\{(6,6)+t\nabla F(6,6): t \in \mathbb{R}\}=\{(6+t,6-t):\ t \in \mathbb{R}\}.$$ The problem is to find some $t \ne 0$ such that $$F(6+t,6-t)=0.$$ Since \begin{eqnarray} F(6+t,6-t)&=&(6+t)^2+3(6+t)(6-t)-4(6-t)^2\\ &=&t^2+12t+36+3(36-t^2)-4(t^2-12t+36)\\ &=&-6t^2+60t=-60t(t-10), \end{eqnarray} we have $$F(6+t,6-t)=0, t\ne 0 \iff t=10,$$ and the corresponding point of intersection is $p'=(16,-4)$.

• Thanks, I like this answer. But one question: why is it $F(6+t,6-t)$? Why add and subtract $t$ from the given points? thanks! Dec 11 '13 at 22:00
• Any point on the normal line at $(6,6)$ is of the form $(6+t,6-t)$ with $t \in \mathbb{R}$. Dec 11 '13 at 22:13
• OK, but why is it of that form? that's all I'm confused about. Thank you. Dec 11 '13 at 22:20
• See the definition of the set $N_p$. Dec 11 '13 at 22:27
• I see your definition of $N_p$, but I don't understand why that is the definition. why is it $(6+t,6-t)$ and not $(6-t,6-t)$ or $(6+t,6+t)$? thank you. Dec 11 '13 at 23:08

Find the equation of the Normal and then solve the two equations, the normal and the curve equations, so you will find the other points of intersection.

You already found the slope of the curve, as a function of $x$ and $y$. So at $(6,6)$ the slope of the curve is

$$y'=\frac{-2x-3y}{3x-8y}=1 \quad \text{when}\,\,x=y=6.$$

The slope of the normal is going to be $-1/y'=-1$.

It remains to define the line passing from $(6,6)$ with slope $-1$.