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.
Best Answer
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 $p'=(16,-4)$.