Question

Locus of a point such that 2 normals drwan from it to parabola are at right angles

Answer 1

Answer:

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).

Step-by-step explanation:

Answer 2

Answer:

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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