Another exam question,

"Find the locus of a the point such that two of the normals drawn through it to the parabola $y^2=4ax$ are perpendicular to each other."

Does the locus mean the point of intersection of the two normals? I attempted to try to this by using the implicit derivative of the parabola and the locus as (x1,y1). Since its given as they meet but I can't get points of intersection.

Can someone help me out please?

  • $\begingroup$ The question is puzzling. In general, from any given point, you can draw one or three normals to a given parabola. See here: demonstrations.wolfram.com/NormalLinesToAParabola $\endgroup$ – bubba Oct 8 '13 at 6:43
  • $\begingroup$ The question is asking you to find all points $p$ such that there are two normals through $p$ to the parabola $y^2=4ax$, and they are perpendicular to each other. $\endgroup$ – Brian M. Scott Oct 8 '13 at 9:08

Animation of the specified locus, for $a = 1/2$:

enter image description here


Find the locus of the point of intersection of two normals to a parabola which are at right angles to one another.


     The equation of the normal to the parabola y^2 = 4ax is 

     y = -tx + 2at + at^3. (t is parameter)

    It passes through the point (h, k) if 

    k = -th + 2at + at^3 => at^3 + t(2a – h) - k = 0.    … (1) 

Let the roots of the above equation be m1, m2and m3. Let the perpendicular normals correspond to the values of m1 and m2 so that m1 m2 = –1.

    From equation (1), m1 m2 m3 = k/a. Since m1 m2 = –1, m3 = -k/a. 

    Since m3 is a root of (1), we have  a(-k/a)^3-k/a (2a – h) - k = 0. 

    ⇒ k^2 = a(h – 3a). 

    Hence the locus of (h, k) is y^2 = a(x – 3a).

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