Equations of normals
Studying for my upcoming exams I've found a topic we didn't even cover in class. Normals.
I've been teaching myself, with help from a classroom chum, but I'm stuck on this question. I posted the answer, the solution, and where it stopped making sense to me. Please help!
First of all, this belongs in the Calculus section.
Second, I don't follow what you don't understand.
You have 2 points that you found a tangent line at.
Now you put those same 2 point in your normal line equation with your new slope and solve to whatever level of simplification or form the problem wants.
Ooh sorry, you're right it's in the wrong section.
I'm not sure where that y-y1=m(x-x1) came from.
Originally Posted by Naur
y-y1=m(x-x1) is the equation of a straight line passing through (x1,y1) and having slope m. see the following page for more equations of straight lines.
in the case of normals m = -1 / (dy/dx) at the point(x1,y1)
Ah I see. So, because we have a point of intersection and the gradient of the line, that formula will give us the equation of its normal...
That's what happens when you teach yourself. You miss out on important basic details. Thanks!
Oh one more thing. Why do they give the answer as x+6y=loge2? Shouldn't it be given with y as the subject, like every other equation?