2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
4
$\begingroup$

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

enter image description here

$\endgroup$
2
$\begingroup$

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

Solution:

     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).
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.