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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?

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    $\begingroup$ Solve the sytem of two equations that you have $\endgroup$ Apr 20 '20 at 20:19
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    $\begingroup$ Simply substitute $y=5x-12$ in the equation of the curve. $\endgroup$ Apr 20 '20 at 20:23
  • $\begingroup$ @MostafaAyaz the equation of the normal to the curve is probably wrong $y=-x+6$ is the normal to the curve $\endgroup$ Apr 20 '20 at 20:39
  • $\begingroup$ @Aryadeva how did you get that equation? $\endgroup$
    – Kari Snow
    Apr 20 '20 at 20:42
  • $\begingroup$ @Aryadeva, yes you are right. The equation must be $y=-x+6$. $\endgroup$ Apr 20 '20 at 20:43
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$$x^2+3xy-4y^2=0$$ 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.

graphing the equation

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  • $\begingroup$ how did you find b? $\endgroup$
    – Kari Snow
    Apr 20 '20 at 20:50
  • $\begingroup$ the normal line goes through the point (3,3) Kari @KariSnow $\endgroup$ Apr 20 '20 at 20:51
  • $\begingroup$ In green it's the normal line @KariSnow $\endgroup$ Apr 20 '20 at 20:54

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