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

Question

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

Using implicit differentiation I got the normal line $y=5x-12$ but now I need to find the other intersecting point. What do I do from here?

Answer 1

$$x^2+3xy-4y^2=0$$ Differentiate: $$2x+3y+3xy'-8yy'=0$$ At $(3,3)$ the slope for the tangent line is: $$15=15y' \implies y'=1$$ $$y=x$$ The normal line is therefore: $$y=-x+b$$ With $(x,y)=(3,3) \implies b=6$ $$y=-x+6$$

Plug this in the original equation to get the intersection points: $$x^2+3xy-4y^2=0$$ $$x^2+3x(-x+6)-4(-x+6)^2=0$$ Solve for x the equation.

Answer 2

It is also possible to exploit the fact that the equation $ \ x^2 + 3xy - 4y^2 \ = \ 0 \ $ describes a "degenerate" conic. Since we can factor the (bi-variate) polynomial as $ \ (x - y)·(x + 4y) \ = \ 0 \ \ , $ this "curve" is actually the union of the two lines $ \ y \ = \ x \ $ and $ \ x \ = \ -4y \ $ (or $ \ y \ = \ -\frac14 x \ ) \ . $ [This is seen in user235711131's graph.] The point $ \ (3 \ , \ 3) \ $ is on the line $ \ y \ = \ x \ \ ; $ since the slope of the line is $ \ m \ = \ 1 \ \ , $ the slope of a normal line to a point on this line is $ \ -\frac{1}{m} \ = \ -1 \ \ . $ Hence, the equation of the normal line at the given point is $ \ y - 3 \ = \ (-1)·(x - 3) \ \ $ or $ \ y \ = \ -x + 6 \ $ (as found in many of the other posts and comments).

Since this line is not parallel to either of the component lines of the degenerate conic, it must intersect each of them. We already have the intersection $ \ (3 \ , \ 3) \ $ on $ \ y \ = \ x \ \ , $ so we want to find the intersection of the normal line with $ \ x \ = \ -4y \ \ . $ We have a choice of ways to do this; one is to solve $$ x \ = \ -4y \ \ \Rightarrow \ \ x \ = \ -4·(-x + 6) \ = \ 4x - 24 \ \ \Rightarrow \ \ 3x \ = \ 24 \ \ \Rightarrow \ \ x \ = \ 8 $$ $$ \Rightarrow \ \ y \ = \ (-8 ) + 6 \ = \ -2 \ \ . $$

Thus, the second intersection point of the chosen normal line with the degenerate conic is at $ \ (8 \ , \ -2) \ \ $ (as also seen in the posted graph).