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

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?

• Solve the sytem of two equations that you have Apr 20 '20 at 20:19
• Simply substitute $y=5x-12$ in the equation of the curve. Apr 20 '20 at 20:23
• @MostafaAyaz the equation of the normal to the curve is probably wrong $y=-x+6$ is the normal to the curve Apr 20 '20 at 20:39
• @Aryadeva how did you get that equation? Apr 20 '20 at 20:42
• @Aryadeva, yes you are right. The equation must be $y=-x+6$. Apr 20 '20 at 20:43

$$x^2+3xy-4y^2=0$$ 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.